I'm trying to prove that, if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.
I proved the existence of $g$ using the Inverse Function Theorem. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,v\in C^1(D)$.
Tha Jacobian is
$\begin{equation*} J =\begin{vmatrix} u_x & u_y\\ v_x & v_y\\ \end{vmatrix} =u_xv_y-u_yv_x \end{equation*}$
and using the Cauchy-Riemann equations, we have that
\begin{equation*} J=(u_x)^2+(v_x)^2=|f'(z)|^2 \end{equation*}
Since $f'(z_0)\neq 0$, $J\neq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.
How can I prove that $g$ is analytic in $w_0$?
That's the only thing that is missing, since I also proved that \begin{equation*} g'(w_0)=\frac{1}{f'(z_0)} \end{equation*}
and since $f'(z_0)\neq 0$, then $g'(w_0)\neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$.