Restriction of a differentiable map $R^3\rightarrow R^3$ to a regular surface is also differentiable.

Restriction of a differentiable map $R^3\rightarrow R^3$ to a regular
surface is also differentiable.

This is again an excercise from Do Carmo's book.
Prove: if $f:R^3 \rightarrow R^3$ is a linear map and $S \subset R^3$ is a
regular surface invariant under $L,$ i.e, $L(S)\subset S$, then the
restriction $L|S$ is a differentiable map and $$dL_p(w)=L(w), p\in S,w\in
T_p(S).$$
My attempt: Since any linear map on $R^3$ can be represented by a linear
transformation matrix , it must be differentiable. By definition I have to
show that for any local parametrization of S say $(U,x)$, map defined by
$x^{-1}\circ L \circ x:U\rightarrow U $ is differentiable locally. Now,
both $x$ and $L$ are differentiable , however , $x^{-1}$ is not
necessarily differentiable.
Moreover, example 3, page 74 of Do Carmo's says : Let $S_1$ and $S_2$ be
regular surfaces. Assume that $S_1\subset V \subset R^3$ where $V$ is an
open subset of $R^3$, and that $\phi:V \rightarrow R^3$ is a
differentiable map such that $\phi(S_1)\subset S_2$. Then the restriction
$\phi|S_1: S_1\rightarrow S_2$ is a differentiable map.
This fact is left without proof, but I think it might be useful for the
question.
Can anyone give me some help ? Thanks in advance