= Solution
Differentiate the position vector, keeping the other coordinates fixed in each <derivative>:
$$
\boxed{\mathbf h_u=(a\sinh u\cos v,a\cosh u\sin v,0),\quad\mathbf h_v=(-a\cosh u\sin v,a\sinh u\cos v,0),\quad\mathbf h_z=(0,0,1).}
$$
The first two have zero <dot product>, since their two planar contributions cancel, and both are orthogonal to $\mathbf h_z$. Hence these are <orthogonal coordinates> away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, are
$$
\boxed{h_u=h_v=a\sqrt{\sinh^2u+\sin^2v},\qquad h_z=1.}
$$
For example $|\mathbf h_u|^2=a^2[\sinh^2u\cos^2v+\cosh^2u\sin^2v]=a^2(\sinh^2u+\sin^2v)$. The unit basis vectors are $\mathbf h_i/h_i$ when $h_i\ne0$; the given coordinate basis vectors themselves are not normalized. The <Jacobian determinant> is $a^2(\sinh^2u+\sin^2v)>0$ except on the focal lines. These are <elliptic cylindrical coordinates>.
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