= Solution
Suppress the held-fixed subscripts only within this matrix display. Collecting the preceding operator coefficients gives
$$
\boxed{\begin{pmatrix}\partial_W\\\partial_\theta\\\partial_\phi\end{pmatrix}
=\begin{pmatrix}
\tan\theta\cos\phi&\tan\theta\sin\phi&1\\
W\sec^2\theta\cos\phi&W\sec^2\theta\sin\phi&0\\
-W\tan\theta\sin\phi&W\tan\theta\cos\phi&0
\end{pmatrix}
\begin{pmatrix}\partial_x\\\partial_y\\\partial_z\end{pmatrix}.}
$$
If $J=\partial(x,y,z)/\partial(W,\theta,\phi)$ uses the usual convention of coordinate <functions> as rows, the displayed coefficient matrix is $J^{\mathsf T}$. Its <determinant> is $W^2\sec^2\theta\tan\theta>0$ on the given domain, so the inverse operator transformation exists everywhere there.
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