Question
$f: \mathbb{R^2} \to \mathbb{R^2} $ is smooth (i.e. $\in C^{\infty}$) and satisfies the Cauchy-Riemann equations i.e. $\frac{\partial f_1} {\partial x}=\frac{\partial f_2} {\partial y}$ and $\frac{\partial f_1} {\partial y}=-\frac{\partial f_2} {\partial x}$.
If $f$ is locally invertible at $x_0$, then $Df(x_0)\neq0$.
I want to solve this question without complex analysis.
I think since locally invertible, there exists $g(y)$ such that $g(f(x))=x$ in a neighborhood of $x_0$.
But I can't show that local invertiblity implies local inverse is differentiable.
If I know that, I think I can show $Df(x_0)\neq0$ by differentiating $g(f(x))=x$.
Help me!!