 Solution

A conservative vector field can be written as $F=\nabla \Phi$ for a single-valued scalar potential $\Phi$. Its line integral between two points depends only on the endpoints, and its integral around every closed curve is zero. A necessary local test for a smooth planar field $F=(P,Q)$ is $P_y=Q_x$. On a simply connected domain such as the whole plane, this is also sufficient.

HereAn explicit scalar potential is even more decisive:Indeed ${\Phi }_x=2x{e}^{xy}+x^{2}y{e}^{xy}$ and ${\Phi }_y=x^3{e}^{xy}+2y$. Thus the field is conservative on ${\mathbb{R}}^2$.

An exact differential is a differential one-form equal to $d\Phi$ on its domain; an inexact differential is not the differential of a single-valued potential there. For example, $d(xy)=ydx+xdy$ is exact. For the supplied form $\eta =42ydx+xdy$, the cross derivatives areThey disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise line integral around the unit square is $1-42=-41\ne 0$. ThereforeThe connection is $F\cdot dr=d\Phi$: a vector field is conservative precisely when its work differential one-form is exact. Locally vanishing curl alone need not guarantee global exactness on a domain with holes.