Prove that the mapping $F(r,\vp,\theta)=(r\cos\vp\cos\theta,r\sin\vp\cos\theta,r\sin\theta)$ is not a conformal mapping from $\R^+\times(-\pi,\pi)\times(-\pi/2,\pi/2)$ into $\R^3$.
Since
$$
DF(r,\vp,\theta)
=\left(\begin{array}{ccc}
\cos\vp\cos\theta&-r\sin\vp\cos\theta&-r\cos\vp\sin\theta\\
\sin\vp\cos\theta&r\cos\vp\cos\theta&-r\sin\vp\sin\theta\\
\sin\theta&0&r\cos\theta
\end{array}\right)
$$
we get
$$
DF(r,\vp,\theta)^*DF(r,\vp,\theta)
=\left(\begin{array}{ccc}
1&0&0\\
0&r^2\cos^2\theta&0\\
0&0&r^2
\end{array}\right)
$$
which evidently is not of the form
$$
h(r,\vp,\theta)^2
\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right)
$$
and thus $F$ is not conformal.