 Solution

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.