= Solution
Differentiate the position <vector> with respect to the three coordinates to obtain the <coordinate basis> vectors:
$$
\boxed{\mathbf h_u=\begin{pmatrix}v\cos\phi\\v\sin\phi\\u\end{pmatrix},\qquad\mathbf h_v=\begin{pmatrix}u\cos\phi\\u\sin\phi\\-v\end{pmatrix},\qquad\mathbf h_\phi=\begin{pmatrix}-uv\sin\phi\\uv\cos\phi\\0\end{pmatrix}.}
$$
The <chain rule> gives $d\mathbf x=\mathbf h_u du+\mathbf h_v dv+\mathbf h_\phi d\phi$. The <scale factors of orthogonal coordinates> are the corresponding <norms>:
$$
\boxed{h_u=h_v=\sqrt{u^2+v^2},\qquad h_\phi=uv.}
$$
Direct <dot products> give $\mathbf h_u\cdot\mathbf h_v=uv-uv=0$, $\mathbf h_u\cdot\mathbf h_\phi=0$, and $\mathbf h_v\cdot\mathbf h_\phi=0$. Thus the <parabolic coordinates> are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.
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