= Solution
The field $\mathbf F$ is not a <conservative vector field>: either nonzero closed <line integral> computed below rules out a <scalar potential>. Equivalently, its planar <curl> is $\partial_xF_y-\partial_yF_x=1-(-1)=2$.
For $\mathbf G$, integrating $\Phi_x=2xy^2$ gives $\Phi=x^2y^2+h(y)$. Matching $\Phi_y=2yx^2$ then gives $h'=0$. Thus
$$
\boxed{\mathbf F\text{ is not conservative};\qquad\mathbf G=\nabla\Phi,\quad\Phi=x^2y^2+C.}
$$
This also explains why both closed <line integrals> of $\mathbf G$ vanish.
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