My theorem states that I need to show the equivalence of 2 ways of having a differentiable function $f:S \longrightarrow \mathbb{R}$ for a point $p \in S$, where $S$ is a regular surface:
a) The differentiability of the composition $f \circ g $ with a parametrization $\phi$ in a neighborhood $W$ of $p$ in $S$.
b) The differentiability of a differentiable extension $F$ of $f$ in a neighborhood of $p$ in $\mathbb{R}^3$.
The proof b) implies a) is really simple by just restricting the domain of the function F.
But I have problem with the opposite proof, a) implies b). Here is my attempt:
If you define a function $\psi: U\subset \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ such that $\psi(x,y,z)=\phi^{-1}(x,y,z)$ where [$\phi^{-1}(x,y,z)=(u(x,y,z), v(x,y,z))$](it takes the same image from $\phi^{-1}$ but it has a different domain) and it follows that $W \subset U$ (U could be a cylinder).
Then one can define $F: U\subset \mathbb{R}^3 \longrightarrow \mathbb{R}$ such that $F=(f \circ \phi) \circ \psi$, and because $f\circ \phi$ is differentiable by hypothesis, and I think $\psi$ is also differentiable, then $F$ is differentiable. Where $F$ is a extension of $f$ because if $F|_{W}=(f \circ \phi) \circ \psi|_{W}=(f \circ \phi) \circ \phi^{-1}=f$.
Do you think that this proof is correct or have you any idea of how to prove it?