Let $p$ be a prime and consider the field $\mathbb{F}\_p$. Fix $f\in\mathbb{F}\_p[X]$ a polynomial of degree $d\ge 2$. Define
$$
K(x,y)=\frac{1}{\sqrt{p}}\sum\_{z\in\mathbb{F}\_p}e\_p(xz+yf(z)),
$$
where $e\_p(t):=e^{2\pi i \frac{t}{p}}$.


It is well-known that if $p\nmid d$, then $|K|$ is bounded by $1$ (up to some constant depending on $d$).


Now consider a sum of products for fixed $t\neq 0$ and $y\neq y'$:
$$
I=\sum\_{x\in\mathbb{F}\_p}K(x,y)\overline{K(x-t,y+t)}\overline{K(x,y')}K(x-t,y'+t).
$$
My question is: for what $f\in\mathbb{F}\_p[X]$ can we bound $|I|$ by $\sqrt{p}$ (up to a multiplicative constant, of course), or perhaps a little worse: $p^{\delta}$ ($\delta<1$).
When $f$ is quadratic, then $K$ is a Gauss sum that can be evaluated explicitly, and I can calculate $I$ and answer my question affirmatively. I don't know what happens for general $f$.



estimate for a sum of products of Weil's sum
Here is a different approach than Denis Chaperon de Lauzières's.


Opening everything and using orthogonality of characters to remove the $x$ variable, we see that your sum is $1/p$ times $$\sum\_{z\_1,z\_2,z\_3,z\_4 \in \mathbb F\_p, z\_1-z\_2-z\_3+z\_4 =0 } e\_p \left( y f(z\_1) + t z\_2 - (y+t) f(z\_2) - y' f(z\_3) -t z\_4 + (y'+t) f(z\_4) \right)$$


This is an exponential sum of a $3$-variable polynomial. Deligne's theorem (Theorem 8.4 of Weil I) shows that this is $O(p^{3/2})$ (with explicit constant $(d-1)^{3}$) as long as $d$ is prime to $p$ and the leading term of the polynomial inside the $e\_p$ defines a smooth hypersurface.


The leading term looks like $$y z\_1^d - (y+t) z\_2^d - y' z\_3^d + (y'+ t) z\_4 ^d $$


The intersection of this with $z\_1 - z\_2 - z\_3 + z\_4 =0$ is smooth unless $$ y^{-1/(d-1)} - (y+t)^{-1/(d-1)} - y' ^{-1/(d-1)} + (y'+t)^{1/d-1}=0$$. 


So you get square-root cancellation outside this set for arbitrary $f$. 


