I'm trying to prove that given two surfaces $S_1$ and $S_2$ and an isometric map $f:S_1\to S_2$, the maps $df_p:T_pS_1\to T_pS_2$ preserve inner products for any $p\in S_1$, i.e., $$\langle df_p(x),df_p(y)\rangle_{T_pS_2}=\langle x,y\rangle_{T_pS_1}$$ for any two $x,y\in T_pS_1$. I don't know very well where to start from, and there is not too much information of this results in many of the classical books (they seem to assume this result as trivial, but it is not trivial at all for me).
By the way, the definition I have of isometric maps is:
A diffeomorphism $f:S_1\to S_2$ is said to be isometric if the length of any curve $\gamma$ in $S_1$ is the same length of the curve $f\circ\gamma$
Pick any $v\in T_pS_1$, let $\gamma : (-\epsilon, \epsilon) \to S_1$ be a smooth curve so that $\gamma'(0) = v$. Then $\gamma_t$ be the restriction of $\gamma$ to $[0,t]$. Then the length of $\gamma_t$ is given by
$$L(\gamma_t) = \int_0^t |\gamma'(s)|ds.$$
On the other hand, using the definition of an isometry,
$$\int_0^t |\gamma'(s)|ds = L(\gamma_t) = L(f\circ \gamma_t) = \int_0^t |df_{\gamma(s)} \gamma'(s)| ds.$$
As the equality is true for all $t$ and $|\gamma'(s)|, |df_{\gamma(s)} \gamma'(s)|$ are both continuous function, we have $$|v| = |\gamma'(0)| = |df_{\gamma(0)} \gamma'(0)| = |df_p (v)|\Rightarrow \langle v, v\rangle = \langle df_p (v), df_p (v)\rangle.$$ Now the general formula follows as the inner product is symmetric.