How can I calculate the inverse of $$ \begin{align*} G: \mathbb{R}^2 &\rightarrow \mathbb{R}^2\\ (x,y) &\mapsto (x, 1-ye^x+xe^y) \end{align*} $$ I actually just need the inverse in a neighborhood of the point $(0,1)$, and I know that the inverse exists in that neighborhood because $$ [D_pG] = \left(\begin{matrix} 1 & 0\\ -ye^x+e^y & xe^y-e^x \end{matrix}\right)_p \implies [D_{(0,1)}G] = \left(\begin{matrix} 1 & 0\\ e-1 & -1 \end{matrix}\right) $$ which is invertible, and by the inverse function theorem I know that there is $G^{-1}:U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^2$ where $G(0,1) \in U$ open set.
Here's what I was able to achieve:
One thing to note is that when $v \rightarrow 0$ we have that
$$ G^{-1}(v) \sim D_{(0,0)}G^{-1}(v) + G^{-1}(0,0) $$
But $G^{-1}(0,0) = G^{-1}(G(0,1)) = (0,1)$ and $D_{(0,0)}G^{-1} = D_{G(0,1)}G^{-1} = (D_{(0,1)}G)^{-1}$, therefore we can conclude that when $v \rightarrow 0$ it follows that
$$ G^{-1}((v_1,v_2)) \sim (D_{(0,1)}G)^{-1}((v_1,v_2)) + (0,1) = (v_1,-v_1-v_2+v_1e+1) $$
Can someone please check my work and give me some directions?
Thanks!