= Solution
Choose the upward orientation of the paraboloid cap. Its boundary is the <circle> $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By <Stokes theorem>,
$$
\int_S(\nabla\times A)\cdot dS=\oint_{\partial S}A\cdot dr.
$$
On the boundary, $dz=0$ and $A\cdot dr=y\,dx-3xy\,dy$. Parametrize by $x=2\cos\theta$, $y=2\sin\theta$, $0\leq\theta\leq2\pi$. Then
$$
\oint_{\partial S}A\cdot dr
=\int_0^{2\pi}[-4\sin^2\theta-24\cos^2\theta\sin\theta]d\theta
=\boxed{-4\pi}.
$$
The second term integrates to zero over a full <period>, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla\times A)_z=-3y-1$, whose integral is $-4\pi$ by symmetry.
An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives \b[$4\pi$].
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