Proof that angle-preserving map is conformal Let $\phi: S \to \bar{S}$ be a diffeomorphism between two surfaces in $\mathbb{R^3}$. Such a map is called conformal if for all $p \in S$, and $v_1, v_2 \in T_p(S)$ (the tangent plane) we have
$$\langle d\phi_p(v_1), d\phi_p(v_2) \rangle = \lambda^2 \langle v_1, v_2 \rangle_p$$
for some nowhere-zero function $\lambda$.
$\phi$ is said to be angle-preserving, if
$$\cos(v_1, v_2) = \cos(d\phi_p(v_1), d\phi_p(v_2)),$$
which I take to mean
$$\frac{\langle v_1, v_2\rangle}{\lVert v_1 \rVert \lVert v_2 \rVert} = 
\frac{\langle d\phi(v_1), d\phi(v_2)\rangle}{\lVert d\phi(v_1) \rVert \lVert d\phi(v_2) \rVert}
$$
From do Carmo, "Differential Geometry of Curves and Surfaces", 4.2/14:

Prove that $\phi$ is locally conformal if and only if it preserves angles.

The "only if" part is obvious, but how can the "if" portion be proved (i.e. how does preserving angles imply conformality)?
 A: Let $e_1$, $e_2$ be an orthonormal basis of $T_{p}S$. Let:
\begin{align*}
\langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \\
\langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \mu \\
\langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_2
\end{align*}
Now take:
\begin{align*}
v_1 &= e_1 \\
v_2 &= \cos\theta\ e_1 + \sin\theta\ e_2
\end{align*}
The equation in your question implies that:
$$
\cos\theta = \frac{\lambda_1 \cos\theta + \mu \sin\theta}{\sqrt{\lambda_1\left(\lambda_1\cos^2\theta + 2\mu\sin\theta\cos\theta + \lambda_2\sin^2\theta\right)}}
$$
Take $\theta = \frac{\pi}{2}$ to get $\mu = 0$. This implies that:
$$
\lambda_1 = \lambda_1 \cos^2\theta + \lambda_2\sin^2\theta
$$
Or $\lambda_1 = \lambda_2$. Hence:
\begin{align*}
\langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \langle e_1, e_1 \rangle_{p} \\
\langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_2, e_2 \rangle_{p} \\
\langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_1, e_2 \rangle_{p} \qquad (= 0)
\end{align*}
Since both $\langle, \rangle_{p}$ and $\langle d\phi_{p}(), d\phi_{p}() \rangle$ are bilinear forms, the above is true for all $v_1, v_2 \in T_{p}S$.
A: Since $d\phi$ is a linear map between 2d spaces, the problem boils down to linear algebra. Given a linear map $T\colon \mathbb R^2\to\mathbb R^2$ that preserves angles between vectors, we would like to show that $T$ is a composition of a unitary with dilation. 
The most efficient approach may depend on your linear algebra background. My weapon of choice is complex numbers: any linear map $T\colon \mathbb R^2\to\mathbb R^2$ is of the form $z\mapsto az +b\bar z$. The angle-preserving assumption means that $\arg (az+b\bar z)=\phi_0\pm \arg z$ where $\phi_0$ is fixed and the sign in $\pm$ is the same for all $z$. So, either $\arg \frac{az+b\bar z}{z}$ or $\arg \frac{az+b\bar z}{\bar z}$ is constant. In the first case we have $b=0$, in the second $a=0$.
