= Solution
Choose the upward orientation. The boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise from above. By <Stokes theorem>, the surface <integral> is $\oint_{\partial S}G\cdot dr$. On this boundary $dz=0$ and $z^2=1$, so
$$
\oint_{\partial S}G\cdot dr=\oint(dx-3xy\,dy).
$$
The first term integrates to zero around a closed curve. By <Green's theorem>, the second equals the disk integral $\int_{x^2+y^2\leq4}(-3y)dx\,dy=0$, by oddness in $y$. Thus
$$
\boxed{\int_S(\nabla\times G)\cdot n\,dS=0.}
$$
Here reversing orientation leaves the zero value unchanged.
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