= Solution
For an independent inversion, use the coordinate relations
$$
W=z,\qquad \theta=\arctan\frac{\sqrt{x^2+y^2}}z,\qquad
\phi=\arctan\frac yx.
$$
The positive-octant domain fixes their angle branches unambiguously. Let $s=\sqrt{x^2+y^2}=W\tan\theta$. Differentiating at fixed Cartesian coordinates gives
$$
\theta_x=\frac{z x}{s(z^2+s^2)}=\frac{\cos^2\theta\cos\phi}{W},\qquad
\theta_y=\frac{\cos^2\theta\sin\phi}{W},\qquad
\theta_z=-\frac{\sin\theta\cos\theta}{W},
$$
and $\phi_x=-y/s^2=-\sin\phi/(W\tan\theta)$, $\phi_y=x/s^2=\cos\phi/(W\tan\theta)$, $\phi_z=0$. Hence the inverse <chain rule> gives
$$
\boxed{(\partial_x)_{y,z}
=\frac{\cos^2\theta\cos\phi}{W}(\partial_\theta)_{W,\phi}
-\frac{\sin\phi}{W\tan\theta}(\partial_\phi)_{W,\theta},}
$$
$$
\boxed{(\partial_y)_{x,z}
=\frac{\cos^2\theta\sin\phi}{W}(\partial_\theta)_{W,\phi}
+\frac{\cos\phi}{W\tan\theta}(\partial_\phi)_{W,\theta},}
$$
and
$$
\boxed{(\partial_z)_{x,y}=(\partial_W)_{\theta,\phi}
-\frac{\sin\theta\cos\theta}{W}(\partial_\theta)_{W,\phi}.}
$$
Multiplying these coefficients by the forward coefficient matrix gives the identity. The original PDF correctly holds $x,y$ fixed in the final <derivative>; the transcription's $x,z$ subscript there is a typo.
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