Let $\mathcal{M}\subset\mathbb{R}^3$ be a smooth enough regular surface. We want to show that around a point $p\in\mathcal{M}$ there is a neighborhood about $p$ in $\mathcal{M}$ which is parametrized conformally.
I don't know if such a proof is possible with my small understanding of differential geometry and complex analysis, but my start was to say that without loss of generality we may assume that a neighborhood of $p\in\mathcal{M}$ is parametrized with $\mathbf{x}:U\to\mathcal{M}$ which is given by $\mathbf{x}=<u,v,f(u,v)>$. So in $R^3$ we can rotate our surface until it is sitting above the $uv$-plane and then we can tediously construct $f$ by drawing lines downward. But, we can also rotate $\mathcal{M}$ to ensure that $T_p\mathcal{M}$ is parallel the $uv$-plane.
So now I am not sure what to do. My idea is to pre(/post?) compose our $\mathbf{x}$ with some functions, write out what makes this composition a conformal mapping, and then we get some differential equations which we can smash with our differential equations hammers. I've had back luck starting this approach though. Any words of advice or thoughts? Thanks!