I would appreciate a feedback on my solution of this problem:
Let $h: \mathbb R\to \mathbb R$ be a $C^1$ map with $h(0)=0$, and consider the system $$e^x+h(y)=u^2, \\ e^y-h(x)=v^2.$$
Show that there is a neighborhood $V\subset\mathbb R^2$ of $(1,1)$ such that for each $(u,v)\in V$ there is a solution $(x,y)$ to this system.
Obviously I need to apply the Implicit function theorem. Consider the map $$f:\mathbb R^{2+2}\to\mathbb R^2,\ (x,y;u,v)\mapsto(e^x+h(y)-u^2, e^y-h(x)-v^2).$$ Observe that $f(0,0;1,1)=(0,0)$. To apply the IFT, I only need to show that the Jacobian w.r.t. the first two variables is nonzero at $(0,0;1,1)$. But the corresponding matrix is \begin{bmatrix}e^x&h'(y)\\-h'(x)&e^y\end{bmatrix} whose determinant at $(0,0;1,1)$ is $1+h'(0)^2 > 0$. Thus by the IFT there is a neighborhood $V$ of $(1,1)$ such that for all $(u,v)\in V$, $f(x,y;u,v)=(0,0)$, i.e., the system has a solution $(x,y)$.