Let $U, V \in \Bbb R^2$ open sets, and $f:U \rightarrow V$ be a bijective $C^1$ smooth function. Denote $f(x,y) = (f_1(x,y), f_2(x,y))$.
Assume: $\forall_{p \in U} \\ \frac{\partial f_1}{\partial x}(p) \neq \frac{\partial f_2}{\partial x}(p) \\ \frac{\partial f_1}{\partial x}(p) + \frac{\partial f_1}{\partial y}(p) = 1 \\ \frac{\partial f_2}{\partial x}(p) + \frac{\partial f_2}{\partial y}(p) = 1 $
Objectives:
$(a)$ Show that the inverse $g=f^{-1}$ is $C^1$ smooth.
$(b)$ Denote $g(u,v) = (g_1(u,v), g_2(u,v))$.
Show $\forall_{q \in V} \\ \frac{\partial g_1}{\partial u}(q) + \frac{\partial g_1}{\partial v}(q) = 1 \\ \frac{\partial g_2}{\partial u}(q) + \frac{\partial g_2}{\partial v}(q) = 1 $
I think both clauses can be solved using the chain rule. I've managed to solve $(a)$ by showing $rk(D_f)=2$ and applying the chain rule to get $rk(D_g)=2$. But I'm not sure how to get $(b)$. Any clues?
Posting the answer for future reference.
Following @Martín-BlasPérezPinilla's tip and applying the chain rule:
$(x,y) = Id_(x,y) = g(f(x,y))$
Taking the derivative of both sides: $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = D_{Id}(x,y) = D_g(f(x,y))D_f(x,y) = D_g(u,v)D_f(x,y)$
By assumption, we have: $D_f(x,y) \begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix}$. Applying D_g(u,v) on both sides yields: $\begin{bmatrix}1\\1\end{bmatrix} = D_g(u,v) \begin{bmatrix}1\\1\end{bmatrix}$.