Let $M \subset \Bbb R^3$ be a regular surface, and ${\bf p} \in M$. Let ${\bf x} : U \subset \Bbb R^2 \to M$ and $\overline{{\bf x}}:\overline{U} \subset \Bbb R^2 \to M$ be parametrizations at ${\bf p}$. Here $U, \overline{U}$ are open (and $\overline{U}$ has nothing to do with the closure of $U$, it is just notation to guide myself here, the less letters, better). If $W = {\bf x}(U) \cap \overline{{\bf x}}(\overline{U}) \ni {\bf p}$, then ${\bf x}^{-1}\circ \overline{{\bf x}}:\overline{{\bf x}}^{-1}(W) \to {\bf x}^{-1}(W)$ is differentiable.
I want to prove this last assertion. Here's my work:
Call ${\bf x}(u,v) = (x(u,v),y(u,v),z(u,v))$ and $\overline{{\bf x}}(\overline{u},\overline{v}) = (x(\overline{u},\bar{v}),y(\bar{u},\bar{v}),z(\bar{u},\bar{v}))$. Fix $(\bar{u}_0,\bar{v}_0) \in \overline{{\bf x}}^{-1}(W)$, and write $(u_0,v_0) = {\bf x}^{-1}\circ \overline{{\bf x}}(\bar{u}_0,\bar{v}_0)$.
Since ${\bf x}$ is regular, suppose WLOG that: $$\frac{\partial(x,y)}{\partial(u,v)}(u_0,v_0) \neq 0.$$ Consider $\varphi: {\bf x}^{-1}(W) \subset \Bbb R^2 \to \Bbb R^2$ given by $\varphi(u,v) = (x(u,v),y(u,v))$. Then: $$\det({\rm d}\varphi_{(u_0,v_0)}) = \frac{\partial(x,y)}{\partial(u,v)}(u_0,v_0) \neq 0,$$ and by the inverse function theorem, exists $V \subset {\bf x}^{-1}(W)$ open containing $(u_0,v_0)$, such that $\varphi^{-1}: \varphi(V) \to V$ exists and is differentiable. We can write: $$\varphi^{-1}(x,y) = (u(x,y),v(x,y)).$$
Let $\pi: \Bbb R^3 \to \Bbb R^2$ be the projection in the first two components. It is differentiable. Hence:
$$\begin{align*} \varphi^{-1} \circ \pi \circ \overline{{\bf x}}(\bar{u},\bar{v}) &= \varphi^{-1}\circ \pi ((x(\overline{u},\bar{v}),y(\bar{u},\bar{v}),z(\bar{u},\bar{v}))) \\ \varphi^{-1} \circ \pi \circ \overline{{\bf x}}(\bar{u},\bar{v}) &= \varphi^{-1} ((x(\overline{u},\bar{v}),y(\bar{u},\bar{v})) \\ \varphi^{-1} \circ \pi \circ \overline{{\bf x}}(\bar{u},\bar{v}) &= (u(x(\overline{u},\bar{v}),y(\bar{u},\bar{v})),v(x(\overline{u},\bar{v}),y(\bar{u},\bar{v}))) \\ \varphi^{-1} \circ \pi \circ \overline{{\bf x}}(\bar{u},\bar{v}) &= {\bf x}^{-1}\circ \overline{\bf x}(\bar{u},\bar{v}) \end{align*} $$
This way, $\varphi^{-1} \circ \pi \circ \overline{{\bf x}}= {\bf x}^{-1}\circ \overline{\bf x}$, and the result follows because the change of parameters is a composition of differentiable functions. So it is differentiable in a neighboorhood of $(\bar{u}_0,\bar{v}_0)$. But even this point was arbitrary.
I'm not used to using the inverse and implicit function theorems, so I appreciate any feedback you guys have. Please see if there is anything wrong with my proof. This is the main issue. If there is nothing wrong, I also welcome easier, alternative proofs. I know that I can just open some books and go after it, but I think it is more instructive to see comments and remarks from people here (and it might be helpful for anyone reading this, also).
Thanks!
You used the same name for the coordinate functions of both $\bf\bar x$ and $\bf x$, so the rest of the proof is not clear for me. Try to fix it, I will edit this answer after that. :)
Since you speak portuguese, you may take a look at theorem 3.1 at this text: http://www.ime.usp.br/~vinior/texts/geodif.pdf