= Solution
Set $s=x^2+y^2>0$. The first two components of the <curl> vanish since the horizontal components are independent of $z$ and the last component depends only on $z$. For the third,
$$
\partial_x\left(\frac{x}{s}\right)=\frac{y^2-x^2}{s^2},\qquad\partial_y\left(\frac{-y}{s}\right)=\frac{y^2-x^2}{s^2},
$$
so \b[this is a <curl-free vector field> everywhere in its domain]:
$$
\boxed{\nabla\times\mathbf F=0.}
$$
Choose the orientation in which the projection onto the $xy$-plane runs counterclockwise, and parametrize
$$
\mathbf x(t)=(\cos t,\sin t,200+\cos t),\qquad0\leq t\leq2\pi.
$$
On this curve the horizontal contribution is $(-\sin t,\cos t)\cdot(-\sin t,\cos t)\,dt=dt$, while the vertical contribution is $z\,dz=d(z^2/2)$. Its integral over a closed curve is zero. \b[Therefore]
$$
\boxed{\oint_C\mathbf F\cdot d\mathbf x=2\pi,}
$$
with $-2\pi$ for the opposite orientation. This does not contradict <Stokes theorem>: the planar disk spanning this curve meets the excluded $z$-axis at $(0,0,200)$, where the <vector field> is undefined. The horizontal one-form is the angular differential and records one winding about that axis; thus this <curl-free vector field> has no single-valued global <potential of a conservative vector field> on its domain.
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