I have the next definition of a conformal map:
A diffeomorphism $f:S_1\to S_2$ between two surfaces is said to be conformal if for any two curves $\gamma_1$ and $\gamma_2$ in $S_1$ intersecting in a point $P$, the angle between the two curves in the point $P$ is equal to the angle between the the curves $f\circ\gamma_1$ y $f\circ\gamma_2$ in the point $f(P)$.
I'm trying to prove that every isometric diffeomorphism is conformal. What I want to show is that $$\frac{f(x)\cdot f(y)}{\|f(x)\|\|f(y)\|}=\frac{x\cdot y}{\|x\|\|y\|}$$ but I don't know where to start from. So I would apreciate some hints.
Let, $S_1$ and $S_2$ be two surfaces with surface patches $\sigma_1$ and $\sigma_2$ respectively.
A diffeomorphism $f:S_1\rightarrow S_2$ is said to be a conformal iff the First fundamental form of $\sigma_1$ of $S_1$ and $\sigma_2=f\circ\sigma_1$ of $S_2$ are proportional.
If $f$ is a local isometry then the First fundamental form of $\sigma_1$ and $\sigma_2$ are the same as the first Fundamental form represents the length of the curve, and hence they are proportional consequently $f$ is conformal.
Hope this works.