= Solution
Substitute the given <Stokes streamfunction> representation into the <divergence> in <spherical coordinates>. The mixed derivatives cancel:
$$
\begin{aligned}
\nabla\cdot\mathbf u
&=\frac1{r^2}\partial_r\left(\frac{\Psi_\theta}{\sin\theta}\right)
+\frac1{r\sin\theta}\partial_\theta\left(-\frac{\Psi_r}{r}\right)\\
&=\frac{\Psi_{r\theta}-\Psi_{\theta r}}{r^2\sin\theta}=0.
\end{aligned}
$$
Thus <incompressibility> is automatic for a smooth <Stokes streamfunction>. Since this is an <axisymmetric flow> with no azimuthal <velocity>, the only possibly nonzero component of its <vorticity> is
$$
\begin{aligned}
(\nabla\times\mathbf u)_\varphi
&=\frac1r\left[\partial_r\left(-\frac{\Psi_r}{\sin\theta}\right)
-\partial_\theta\left(\frac{\Psi_\theta}{r^2\sin\theta}\right)\right]\\
&=-\frac1{r^3\sin\theta}\left[r^2\Psi_{rr}+\sin\theta\,\partial_\theta\left(\frac{\Psi_\theta}{\sin\theta}\right)\right].
\end{aligned}
$$
Define the differential operator in the <Stokes streamfunction equation for irrotational flow>
$$
\mathcal L\Psi=r^2\Psi_{rr}+\Psi_{\theta\theta}-\cot\theta\,\Psi_\theta.
$$
The stated equation is $\mathcal L\Psi=0$, and therefore
$$
\boxed{\nabla\cdot\mathbf u=0,\qquad\nabla\times\mathbf u=\mathbf0.}
$$
The <irrotational flow> has a <velocity potential> locally, and globally on a regular <simply connected domain>. Regularity of the physical <velocity field> supplies the limiting interpretation on the polar axis, where the coordinate expressions contain $\sin\theta$ in the denominator.
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