= Solution
An <exact differential> is a one-form equal to $d\Phi$ for a single-valued scalar function $\Phi$ on the domain. In Euclidean coordinates this means its vector coefficients form $\nabla\Phi$. Integrating the first component of $F$ with respect to $x$, then matching the other components, gives
$$
\boxed{\Phi(x,y,z)=z^3(e^x-e^y)+x^3(e^y-e^z)+C.}
$$
Differentiating checks all three components. Since the domain $\mathbb R^3$ is connected, two such potentials differ only by a constant, so this is the most general potential.
At both specified endpoints the nonconstant terms vanish. The <fundamental theorem for line integrals> gives \b[zero for the line integral along every path joining the two points].
For the planar field, choose a continuous lift $\theta(t)$ of the polar angle along a curve avoiding the origin. Differentiating the angle gives
$$
G\cdot dx=\frac{-y\,dx+x\,dy}{x^2+y^2}=d\theta.
$$
On the closed curve the angle increases by $2\pi$ times its <winding number>. Thus
$$
\boxed{\oint_CG\cdot dx=2\pi.}
$$
The angle differential is locally exact but not globally exact on the <punctured plane>: the angle lift need not return to its initial value. This is the <nonexact angular one-form>, whose nonzero closed-path integral obstructs a global potential.
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