= 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.
Here
$$
P_y=(3x^2+x^3y)e^{xy}=Q_x.
$$
An explicit <scalar potential> is even more decisive:
$$
\boxed{\Phi(x,y)=x^2e^{xy}+y^2,\qquad F=\nabla\Phi.}
$$
Indeed $\Phi_x=2xe^{xy}+x^2ye^{xy}$ and $\Phi_y=x^3e^{xy}+2y$. 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, $d(xy)=y\,dx+x\,dy$ is exact. For the supplied form $\eta=42y\,dx+x\,dy$, the cross derivatives are
$$
\frac{\partial(42y)}{\partial y}=42,\qquad
\frac{\partial x}{\partial x}=1.
$$
They disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise <line integral> around the unit square is $1-42=-41\ne0$. Therefore
$$
\boxed{42y\,dx+x\,dy\text{ is inexact}.}
$$
The connection is $F\cdot d r=d\Phi$: 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.
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