As William mentioned, you can construct your function $g$ if $f$ has a local inverse about the origin. Hence we must use the inverse function theorem. First note that $f$ is continuously differentiable everywhere. Moreover, we can compute the Jacobian $J$:
$J(x,y) = \begin{bmatrix} 1 & 2e^{2y} \\ 2x\cos(x^2+y^2) & \cos(x^2 + y) \end{bmatrix}$.
Note that at $(0,0)$, $\det(J(0,0)) = \left| \begin{matrix} 1 & 2 \\ 0 &1\end{matrix}\right| = 1$. So we have satisfied the hypotheses of the inverse function theorem, thus we can construct such a $g$. But the inverse function theorem is even stronger in that it gives you a formula for the derivative of $f^{-1}$ locally in terms of $f'$. You can probably work it out from there by observing $f^{-1}(0,0) = (0,0)$.