Let $h:R\to R$ be a $C^1$ function, $h(0)=0$. Consider the system $$e^x+h(y)=u^2 \\ e^y-h(x)=v^2$$ Prove that there is a neighborhood $V$ of $(1,1)$ such that for all $(u,v)\in V$ there is a solution $(x,y)$.
Is my solution correct?
Consider $F: (x,y,u,v)\mapsto (e^x+h(y)-u^2,e^y-h(x)-v^2)$. It is $C^1$, with $F(0,0,1,1)=(0,0)$. The Jacobian determinant w.r.t. the first two variables at $(0,0)$ is $1+h'(0)^2$. So there is an open set $V\subset R^2$ containing $(1,1)$ and a $C^1$ function $g=(g_1,g_2):V\to R^2$ such that $g(1,1)=(0,0)$ and $F(g_1(u,v),g_2(u,v),u,v)=(0,0)$ for all $(u,v)\in V$. That is, there is a local solution $(x,y)=(g_1(u,v),g_2(u,v))$.