Suppose $M,N$ are open subsets of $\R^n$. A smooth map $F:M\rar N$ is conformal if and only if it preserves angles.
We already know that conformal mappings preserve angles. Hence it suffices to prove that any smooth mapping $F$ satisfying
$$
\forall x\in M\,\forall u,v\in\R^n:\quad
\frac{\la DF(x)u,DF(x)v\ra}{\Vert DF(x)u\Vert\Vert DF(x)v\Vert}
=\frac{\la u,v\ra}{\Vert u\Vert\Vert v\Vert}
$$
there is some smooth function $h:M\rar\R^+$ such that
$$
\la DF(x)u,DF(x)v\ra=h(x)^2\la u,v\ra~.
$$
Let $e_j$ be an orthonormal basis for $\R^n$, then for $u_j\colon=\la u,e_j\ra$:
$$
\la DF(x)u,DF(x)v\ra=\sum g_{jk}(x)u_jv_k
\quad\mbox{where}\quad
g_{jk}(x)=\la DF(x)e_j,DF(x)e_k\ra,
$$
i.e. $(g_{jk}(x))$ is the Gramian of the Jacobi matrix of $F$ at $x$. By assumption we have for all $j,k$:
$$
\frac{g_{jk}}{\sqrt{g_{jj}g_{kk}}}=\d_{jk}
$$
in particular for $j\neq k$: $g_{jk}=0$, i.e.
$$
\la DF(x)u,DF(x)v\ra=\sum_j g_{jj}(x)u_jv_j
$$
Finally for $u=(e_1+e_j)/\sqrt2$ and $v=e_1$ we get:
$$
1/\sqrt2
=\la u,v\ra
=\frac{\sqrt{g_{11}}}
{\sqrt{g_{11}+g_{jj}}}
$$
which can only hold if $g_{jj}=g_{11}$. If follows that $h=\sqrt{g_{11}}$.