As part of a different problem, I came across the following simplified question, for which I cannot exhibit a proof nor a counterexample. Note that the assumptions of smoothness and strict positivity of $f$ away from the $y$-axis are important. 


Let $f(x,y): \mathbb{R}^2 \rightarrow \mathbb{R}$ be a smooth (${\cal C}^\infty$) function such that $f(0,y)=0$ for all $y \in \mathbb{R}$ and $f(x,y)>0$ for $x \neq 0$. Does the following hold:


There exists $y\_0 \in \mathbb{R}$, $\varepsilon >0$ such that $f(x,y\_0)+\int\_0^x \frac{\partial f}{\partial y}(s,y\_0)ds >0$ for $|x|<\varepsilon, x \neq 0$.



Lower bounds on derivative around zero set of a positive smooth function
Carlos pointed out that this works for analytic functions, leaving the general case open.


I believe, in the general case, this is false. I think something like $e^{-1/x^2} \left(\sin\left( y- \frac{1}{x} \right)+1\right)+g(x)$, for $g(x)$ a sufficiently small positive smooth function, should do. This is smooth, because any factors that occur as we take the derivative will be easily dwarfed by $e^{1/x^2}$. For each value of $y$, $\sin\left( y- \frac{1}{x} \right) =-1$ for infinitely many $x$ near $0$. So no matter what $y\_0$ is or how small $\epsilon$ is, you will always be able to find an $x$ with $f(x,y\_0)=g(x)$ and $x>0$.


Now we integrate $\int\_{0}^{x} e^{-1/s^2} \cos\left(y\_0-\frac{1}{s}\right) ds$, where $y\_0-\frac{1}{s} \equiv -\pi/2$ modulo $2\pi$. We can break the integral up into intervals of the form $\left[\frac{1}{n\pi+\pi/2+y\_0},\frac{1}{n\pi-\pi/2+y\_0}\right]$. Let the integral over this interval be $A\_n$.


As $n$ increases, $e^{-1/s^2}$ goes to $0$, as does the width of the interval, so $|A\_n|$ is a decreasing function of $n$. The sign of $\cos\left( y\_0 -\frac{1}{s}\right)$ does not change. It is always $(-1)^n$, so the sign of $A\_n$ is $(-1)^n$.


Thus


$$\int\_{0}^{x} e^{-1/s^2} \cos\left(y\_0-\frac{1}{s}\right) ds = \sum\_{n=k}^\infty A\_n= (A\_k+A\_{k+1}) + (A\_{k+2} + A\_{k+3})+\dots$$


In each pair, the first one is dominant, and has sign $(-1)^k$, so $\int\_{0}^{x} e^{-1/s^2} \cos\left(y\_0-\frac{1}{s}\right) ds$ has sign $(-1)^k$. Since $y\_0-\frac{1}{s} \equiv -\pi/2$ modulo $2\pi$, $k$ is odd, and this is negative.


Clearly if we did a slightly messy calculation, we could find some uniform lower bound on the size of this integral. Since $f(x,y\_0)=g(x)$ here, we just need to choose a $g(x)$ smooth, positive, and smaller than this bound to find a counterexample.


