Since $f=y'=e^{2y}$ and $\frac{\partial f}{\partial y}=2e^{2y}$ are continuous everywhere in $\mathbb{R}$, any interval of $x$ containing $(0,0)$ will have a unique solution satisfying the IVP.

In that case we should just be able to choose the smallest interval, since all of them work. $|x| \leq \frac{1}{2e}$ in this case.