Solution
= Solution
{c}
Write $r=\sqrt{x^2+y^2}>0$, so $\mathbf F=(f(z)x/r,f(z)y/r,0)$. The Cartesian <curl> is
$$
\nabla\times\mathbf F
=\left(-f'(z)\frac yr,\ f'(z)\frac xr,\
\partial_x\frac{f(z)y}r-\partial_y\frac{f(z)x}r\right).
$$
The last component is zero because both differentiated terms are $-f(z)xy/r^3$. Thus
$$
\boxed{\nabla\times\mathbf F=f'(z)(-\sin\theta,\cos\theta,0)=f'(z)\mathbf e_\theta.}
$$
The condition $r>0$ excludes the cylindrical-axis coordinate singularity.