 Solution

A differential one-form $Pdx+Qdy$ is an exact differential if there is a differentiable function $F$ with $P=F_x$ and $Q=F_y$, so the form equals $dF$. For continuously differentiable $P,Q$, equality of mixed partial derivatives gives the necessary conditionA local differential condition need not by itself give a global potential on a domain with holes; no such issue arises for the explicit primitives here.

For the first form, $P_y=2y=Q_x$ and direct differentiation givesFor the second, $P_y=2y$ but $Q_x=y^2$. These are not equal on any nonempty open subset of the plane, so that form is not an exact differential. Nevertheless,Thus one suitable pair is $f(x,y)=x{e}^y$, $g(y)=y^2{e}^{-y}$. This factorization is smooth even at $y=0$. On regions where $y\ne 0$, its reciprocal factor $e^y/y^2$ is an integrating factor which turns the original form into $df$.