= Solution
{c}
Here $\phi$ is a <scalar field>: the source's description as a <vector field> is a typographical slip, since the displayed formula is for its <gradient>. For <cylindrical coordinates>, write $x=r\cos\theta$, $y=r\sin\theta$. Away from $r=0$, a local choice of $\theta$ gives the <partial derivatives>
$$
r_x=\cos\theta,\quad r_y=\sin\theta,\quad
\theta_x=-\frac{\sin\theta}{r},\quad
\theta_y=\frac{\cos\theta}{r}.
$$
The <chain rule> therefore gives
$$
\phi_x=\Phi_r\cos\theta-\frac{\Phi_\theta}{r}\sin\theta,
\qquad
\phi_y=\Phi_r\sin\theta+\frac{\Phi_\theta}{r}\cos\theta,
\qquad \phi_z=\Phi_z.
$$
Use the <Cartesian coordinates> definition $\nabla\phi=\phi_x\mathbf e_x+\phi_y\mathbf e_y+\phi_z\mathbf e_z$ and the <unit vectors>
$$
\mathbf e_r=\cos\theta\,\mathbf e_x+\sin\theta\,\mathbf e_y,
\qquad
\mathbf e_\theta=-\sin\theta\,\mathbf e_x+\cos\theta\,\mathbf e_y.
$$
Collecting their coefficients proves
$$
\boxed{\nabla\phi=\Phi_r\mathbf e_r+\frac1r\Phi_\theta\mathbf e_\theta+\Phi_z\mathbf e_z,\qquad r>0.}
$$
This is the <gradient in cylindrical coordinates>. The factor $1/r$ converts a change in angle into a change per unit <arc length>.
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