= Solution
In Cartesian components, $\mathbf F=(z-y,x,-x)$, so
$$
\boxed{\nabla\times\mathbf F=(0,2,2)\ne0.}
$$
It 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.
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