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.