= Solution
Write $r^2=x^2+y^2$ and work away from the $z$ axis, where $r>0$. The <vector field> is independent of $z$ and has zero vertical component, so only the vertical <curl> can be nonzero:
$$
\begin{aligned}
(\nabla\times F)_z
&=\frac{\partial}{\partial x}\big[x(x^2+y^2)^{-n}\big]
-\frac{\partial}{\partial y}\big[-y(x^2+y^2)^{-n}\big]\\
&=2r^{-2n}-2n(x^2+y^2)r^{-2n-2}.
\end{aligned}
$$
Therefore
$$
\boxed{\nabla\times F=2(1-n)r^{-2n}\,\hat z,\qquad r>0.}
$$
This <local spin in an azimuthal power-law flow> depends on the radial variation of the tangential speed, not merely on the sign of that speed.
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