Existences of Solutions by Using Rolle's Theorem

The question is:

Let: $$g(x,y)=(e^x+1)y^2+2(e^{x^2}-e^{2x-1})y+(e^{-x^2}-1)$$ Then, show that there exists a constant $$\bar{y}>0$$, such that for any fixed $$y \in [0,\bar{y}]$$, the equation $$g(x,y) = 0$$ admits a solution $$x(y)$$.

This question is with the background of an exercise about Rolle's Theorem. However, it is still possible that the solution isn't unique.

The problem is I have no idea about how to apply it. I have considered first integrate $$g(x, y)$$ with respect to $$x$$, then apply the Rolle's Theorem to the point $$0$$ and $$\bar{y}$$ in the integrated function but I failed because I can't integrate $$e^{x^2}$$ and I failed to find an appropriate $$\bar{y}$$.

Any help is appreciated! Many thanks.

It is obvious that for y>0 :$$\lim_{x\to\infty} g(x,y)=+\infty$$ Then we just need to find a constant $$\overline{y}$$ such that, for $$y\in]0,\overline{y}]$$ , the inequality : $$g(x,y)<0$$ admits a solution x
Let's take $$x=1$$, we have $$g(1,y)=(e+1)y^2+\frac{1}{e}-1$$, so, if we take $$\overline{y}=\sqrt{\frac{1-\frac{1}{e}}{e+1}}$$ then for every $$y\in]0,\overline{y}]$$ : $$g(1,y)<0$$ Thus we can apply the role theorem to the function $$x\mapsto g(x,y)$$ which gives the existence of the solution x(y) for every $$y\in]0,\overline{y}]$$ Finally, for $$y=0$$ , the solution is $$x(y)=0$$ .