= Solution
{c}
Write $r=\sqrt{x^2+y^2}>0$. The <cylindrical coordinates> give $\mathbf e_\theta=(-y/r,x/r,0)$ and $\mathbf e_r=(x/r,y/r,0)$, so the <vector components> of the <vector field> are
$$
F_x=-\frac{yf(z)}r,\qquad F_y=\frac{xf(z)}r,\qquad F_z=0.
$$
The first two components of the <curl>, calculated in <Cartesian coordinates>, are $-xf'(z)/r$ and $-yf'(z)/r$. For the third component, the <chain rule> gives
$$
\partial_x(x/r)=\frac1r-\frac{x^2}{r^3},\qquad
\partial_y(y/r)=\frac1r-\frac{y^2}{r^3},
$$
and hence
$$
(\nabla\times\mathbf F)_z=f(z)\left(\frac2r-\frac{x^2+y^2}{r^3}\right)=\frac{f(z)}r.
$$
Combining these <vector components> in the <cylindrical basis> yields
$$
\boxed{\nabla\times\mathbf F=\frac{f(z)}r\mathbf e_z-f'(z)\mathbf e_r,\qquad r>0.}
$$
The restriction $r>0$ matters because the <cylindrical basis> is undefined on the axis.
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