 Solution

In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.