First order partial derivatives Suppose that $f:\Bbb R^2\to \Bbb R^2$ has $C^1$ partial derivatives in some ball $B_r(x_0,y_0)$ $r>0$. Prove that if $\Delta_f(x_0,y_0)\neq 0$, then
$\displaystyle\frac{\partial f_1^{-1}}{\partial x}(f(x_0,y_0))= \displaystyle\frac{\partial f_2/\partial y(x_0,y_0)}{\Delta_f(x_0,y_0)}$
$\displaystyle\frac{\partial f_1^{-1}}{\partial y}(f(x_0,y_0))= \displaystyle\frac{\partial f_1/\partial y(x_0,y_0)}{\Delta_f(x_0,y_0)}$
$\displaystyle\frac{\partial f_2^{-1}}{\partial x}(f(x_0,y_0))= \displaystyle\frac{\partial f_2/\partial x(x_0,y_0)}{\Delta_f(x_0,y_0)}$
$\displaystyle\frac{\partial f_2^{-1}}{\partial xy}(f(x_0,y_0))= \displaystyle\frac{\partial f_1/\partial x(x_0,y_0)}{\Delta_f(x_0,y_0)}$
Please can someone show me only one equation? And then I Will beraber to solve second equation by myself. Thank you 
 A: You can prove all of them at once. Recall the inverse function theorem: Suppose that $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable in an open set containing $a$ and $\det f'(a) \neq 0 $. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f : V \to W$ has a continuous inverse $f^{-1} : W \to V$ which is differentiable and for all $y \in W$ satisfies:
$$(f^{-1})'(y)=\left[f'(f^{-1}(y))\right]^{-1}$$
Notice that your $f$ is continuously differentiable in $B_r(x_0,y_0)$ (which is an open set) and recall that if $f'(x_0,y_0)$ is the jacobian matrix, then by your hypothesis we have that $\det f'(x_0, y_0) \neq 0$. So your function fits the hypothesis of the theorem. Then there must be an open set $V$ containing $(x_0,y_0)$ and an open set $W$ containing $f(x_0,y_0)$ such that $f : V \to W$ has continuous inverse $f^{-1} :  W \to V$ which is differentiable for all $(a,b) \in W$. And such that the jacobian matrix at each point $(a,b) \in W$ is given by the rule:
$$(f^{-1})'(a,b)=\left[f'(f^{-1}(a,b))\right]^{-1}$$
Take now $(a,b)=f(x_0,y_0)$. Then we have:
$$(f^{-1})'(f(x_0,y_0))=\left[f'(f^{-1}(f(x_0,y_0)))\right]^{-1}$$
But $f^{-1}(f(x_0,y_0)) = (x_0,y_0)$ so that this is the same as:
$$(f^{-1})'(f(x_0,y_0))=\left[f'(x_0,y_0)\right]^{-1}$$
Now can you procede from here?
