= Solution
Compare the <total differential> $d\phi=\sum_i\partial_{u_i}\phi\,du_i$ with $d\phi=\nabla\phi\cdot d\mathbf r$ and $d\mathbf r=\sum_i h_i\mathbf e_i\,du_i$. The orthonormality of the coordinate basis gives the <gradient in orthogonal coordinates>:
$$
\boxed{\nabla\phi=\sum_{i=1}^3\frac{\mathbf e_i}{h_i}\partial_{u_i}\phi,\qquad \nabla u_i=\frac{\mathbf e_i}{h_i}.}
$$
To obtain the <curl> without neglecting the variation of the basis vectors, write $\phi\mathbf e_1=(\phi h_1)\nabla u_1$. The curl of a gradient vanishes, and the product rule gives
$$
\nabla\times(\phi\mathbf e_1)=\nabla(\phi h_1)\times\nabla u_1=\frac{\mathbf e_2}{h_1h_3}\partial_{u_3}(\phi h_1)-\frac{\mathbf e_3}{h_1h_2}\partial_{u_2}(\phi h_1).
$$
Likewise $\nabla u_2\times\nabla u_3=\mathbf e_1/(h_2h_3)$ because the system is right-handed. This cross product has zero <divergence>: $\nabla\cdot(\nabla u_2\times\nabla u_3)=\nabla u_3\cdot(\nabla\times\nabla u_2)-\nabla u_2\cdot(\nabla\times\nabla u_3)=0$. The divergence product rule applied to $(\phi h_2h_3)(\nabla u_2\times\nabla u_3)$ now gives
$$
\nabla\cdot(\phi\mathbf e_1)=\frac1{h_1h_2h_3}\partial_{u_1}(\phi h_2h_3).
$$
Cyclic permutation and linearity give the general component formulas described by <divergence and curl from orthogonal coordinate gradients>:
$$
\boxed{\nabla\cdot\mathbf v=\frac{\partial_{u_1}(h_2h_3v_1)+\partial_{u_2}(h_3h_1v_2)+\partial_{u_3}(h_1h_2v_3)}{h_1h_2h_3},}
$$
and
$$
\boxed{\nabla\times\mathbf v=\frac{\mathbf e_1}{h_2h_3}[\partial_{u_2}(h_3v_3)-\partial_{u_3}(h_2v_2)]+\frac{\mathbf e_2}{h_3h_1}[\partial_{u_3}(h_1v_1)-\partial_{u_1}(h_3v_3)]+\frac{\mathbf e_3}{h_1h_2}[\partial_{u_1}(h_2v_2)-\partial_{u_2}(h_1v_1)].}
$$
In <cylindrical coordinates>, $(h_r,h_\theta,h_z)=(1,r,1)$. The specified field has no $z$ dependence or $z$ component, so
$$
\boxed{\nabla\times\mathbf v=\left[\frac1r\partial_r(r^4)-\frac1r\partial_\theta(\alpha\theta r^p)\right]\mathbf e_z=(4r^2-\alpha r^{p-1})\mathbf e_z.}
$$
It is an <irrotational vector field> for every $r>0$ exactly when $\boxed{p=3,\ \alpha=4}$. This statement is local to an angular coordinate chart: the component proportional to $\theta$ is not single-valued under $\theta\mapsto\theta+2\pi$.
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