Let $0<\theta\_1,\theta\_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.


Consider the oscillatory integral
$$I(t):=\int\_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta\_1})
(y-e^{\dot{\imath}\theta\_2})}\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$
I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.


Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives


$$\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$
with some constant $C$. But then we are faced with the singularities $y^{-1}$ and $y^{-2}$.


If we use Fubini's theorem then apply the stationary phase method we end up with


$$\int\_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta\_1})
(y-e^{\dot{\imath}\theta\_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$
which is again not integrable against $\psi$.


Another approach is to change variables $s=\sqrt{t}r$ to write


$$I(t)=\frac{1}{\sqrt{t}}
J(t)$$
where
$$J(t):=\int\_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta\_1})
(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta\_2})}\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$
The idea now is to check whether $\lim\_{t\rightarrow \infty}J(t)$
is a constant independent of $t$. But, by Fubini's theorem, we have
$$\lim\_{t\rightarrow \infty}J(t)=c\int\_{\mathbb{R}}
\frac{\psi(x)}{x}dx$$ which may diverge.



Asymptotic behavior of a double oscillatory integral
$\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$Obviously, for $a:=\sqrt{2\pi}\,\psi(0)$ we have
\begin{equation\*}
 \psi(0)=f(0), 
\end{equation\*}
where
\begin{equation\*}
 f:=a\vpi
\end{equation\*}
and $\vpi$ is the standard normal density. Letting now $g(x):=\frac{\psi(x)-f(x)}x$ for $x\ne0$, with $g(0):=\psi'(0)$, we see that $g$ is a smooth integrable function and
\begin{equation\*}
 \psi(x)=f(x)+xg(x) \tag{1}\label{1}
\end{equation\*}
for all real $x$.


So,
\begin{equation\*}
 I(t)=\frac{J\_f(t)+J\_h(t)}{\sqrt t}, \tag{2}\label{2}
\end{equation\*}
where $h(x):=xg(x)$,
\begin{equation\*}
J\_f(t):= 
\int\_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)}
\int\_{\R}dx\,f(x)e^{ix^2y^2}, 
\end{equation\*}
\begin{equation\*}
 z\_1:=e^{i\theta\_1},\quad z\_2:=e^{i\theta\_2},
\end{equation\*}
with $J\_h(t)$ defined similarly. Here and in what follows, $t$ is any real number $>0$, unless specified otherwise.


Note that
\begin{equation\*}
 \Im z\_1\ne0\quad\text{and}\quad\Im z\_2\ne0. \tag{3}\label{3}
\end{equation\*}
Note also that
\begin{equation\*}
 \int\_{\R}dx\,f(x)e^{ix^2y^2}=\frac a{\sqrt{1-2 i y^2}}
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
J\_f(t)&= 
a\int\_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}. 
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
For any fixed real $A>0$, by dominated convergence (which holds in view of \eqref{3}),
\begin{equation\*}
 \int\_0^A\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}
 \to\frac1{z\_1z\_2}\int\_0^A\frac{dy}{\sqrt{1-2 i y^2}}\ll1 
\end{equation\*}(as $t\to\infty$).
We write $E\ll F$ to mean $|E|=O(F)$.


So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$ and
\begin{equation\*}
 \int\_0^A\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}
 =o(\ln t). 
\end{equation\*}
Also, we will have
\begin{equation\*}
 \int\_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}} \\ 
 \sim\int\_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)y\sqrt{-2 i}}
\sim\frac{1+i}{4z\_1z\_2}\,\ln t. \tag{5}\label{5}
\end{equation\*}
The latter asymptotic expression in \eqref{5} can be obtained by taking the latter integral in \eqref{5} in closed form, by partial fraction decomposition. It can also be obtained by writing $\int\_A^{\sqrt t}=\int\_A^{t^b}+\int\_{t^b}^{t^{1/2}}$ for $b\in(0,1/2)$ such that $b$ is close to $1/2$.


So, assuming $\psi(0)\ne0$, by \eqref{4},
\begin{equation\*}
J\_f(t)\sim
a\frac{1+i}{4z\_1z\_2}\,\ln t
=\frac{1+i}{4z\_1z\_2}\,\sqrt{2\pi}\,\psi(0)\ln t. 
\tag{6}\label{6}
\end{equation\*}


Next,
\begin{equation\*}
J\_h(t)= \int\_{\R}dx\,\sgn(x)g(x)K(t,x), \tag{7}\label{7}
\end{equation\*}
where
\begin{equation\*}
 K(t,x):=\int\_0^{\sqrt t}\frac{dy\,e^{ix^2y^2}|x|}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)}
 =\frac12\int\_0^{r^2}dv\, e^{iv}H\_r(v), 
\end{equation\*}
\begin{equation\*}
 r:=|x|\sqrt t, 
\end{equation\*}
\begin{equation\*}
 H\_r(v):=\frac1{\sqrt v\,(\frac{\sqrt v}r-z\_1)(\frac{\sqrt v}r-z\_2)}. 
\end{equation\*}
Further,
\begin{equation\*}
 2K(t,x)=K\_1+K\_2, \tag{8}\label{8}
\end{equation\*}
where
\begin{equation\*}
 K\_1:=\int\_0^{1\wedge r^2}dv\, e^{iv}H\_r(v),\quad K\_2:=\int\_{1\wedge r^2}^{r^2}dv\, e^{iv}H\_r(v),
\end{equation\*}
$u\wedge w:=\min(u,w)$.


For $v\in(0,1\wedge r^2)$, we have $H\_r(v)\ll\frac1{\sqrt v}$ and hence
\begin{equation\*}
 K\_1\ll1. \tag{9}\label{9}
\end{equation\*}
Note that $K\_2=0$ if $r\le1$. If now $r>1$, then
\begin{equation\*}
 H'\_r(v)=-\frac{H\_r(v)}{2v}\,\Big(1+\frac1{\sqrt v-rz\_1}+\frac1{\sqrt v-rz\_2}\Big)
 \ll \frac{|H\_r(v)|}v\ll\frac1{v^{3/2}} 
\end{equation\*}
for $v>0$; so, integrating by parts, we see that  

\begin{equation\*}
 K\_2\ll1. \tag{10}\label{10} 
\end{equation\*}


Collecting \eqref{7}--\eqref{10} and recalling that $g$ is an integrable function, we see that
\begin{equation\*}
J\_h(t)\ll1. \tag{11}\label{11}
\end{equation\*}


Collecting \eqref{2}, \eqref{6}, and \eqref{10}, we conclude that
\begin{equation\*}
I(t)\sim
\frac{1+i}{4z\_1z\_2}\,\sqrt{2\pi}\,\psi(0)\frac{\ln t}{\sqrt t}, \tag{12}\label{12}
\end{equation\*}
as $t\to\infty$.




---


As seen from the above proof, for \eqref{12} to hold it is enough that the function $\R\setminus\{0\}\ni x\mapsto\frac{\psi(x)-\psi(0)}x$ be integrable, with $\psi(0)\ne0$. (Of course, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)




---


Concerning the case $\psi(0)=0$: Then the bounds on $H\_r(v)$ and $H'\_r(v)$ developed above provide for dominated convergence. So, taking into account that for each real $x\ne0$
\begin{equation}
 H\_r(v)\to\frac1{z\_1z\_2\sqrt v}, 
\end{equation}
we have
\begin{equation}
 K(t,x)\to\frac1{2z\_1z\_2}\int\_0^\infty\frac{dv\,e^{iv}}{\sqrt v}
 =\frac{1+i}{2z\_1z\_2}\,\sqrt{\frac\pi2}
\end{equation}
and
\begin{equation\*}
J\_h(t)\to C\_\psi:= c\_\psi \frac{1+i}{2z\_1z\_2}\,\sqrt{\frac\pi2}, 
\end{equation\*}
where
\begin{equation}
 c\_\psi:=\int\_{\R}dx\,\sgn(x)g(x)=\int\_{\R}dx\,\frac{\psi(x)}{|x|}. 
\end{equation}
Therefore and because here $f=0$ and hence $J\_f(t)=0$, we conclude that
\begin{equation\*}
I(t)\sim \frac{C\_\psi}{\sqrt t}, 
\end{equation\*}
as $t\to\infty$, provided that $\psi(0)=0$, $c\_\psi\ne0$, and the function $g$ is integrable. (In the case when $\psi(0)=0$, we have $g(x)=\psi(x)/x$ for $x\ne0$. As was already noted, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)


If $\psi(0)=0$ and $c\_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.


However, if $\psi$ is a "smooth bump", as you said in a comment, then apparently $\psi\ge0$ and hence $c\_\psi>0$ (unless $\psi=0$ and hence $I(t)=0$).


