= Solution
Let $F'=f$ and $G'=g$ on the relevant intervals. The <chain rule> gives
$$
d\{F(xy)+G(y/x)\}
=\left[yf(xy)-\frac y{x^2}g(y/x)\right]dx
+\left[xf(xy)+\frac1xg(y/x)\right]dy.
$$
Thus the one-form is explicitly an <exact differential>. Fixing its potential to vanish at $(1,1)$ determines it uniquely on the connected domain:
$$
\boxed{\phi(x,y)=\int_1^{xy}f(s)\,ds+\int_1^{y/x}g(t)\,dt.}
$$
Continuity of $f,g$ on the integration ranges suffices. For arbitrary smooth data defined throughout $x>0$, that half-plane also supplies the simply connected domain in the previous argument.
For the specialization $f(s)=s^{-2}$, the potential based at $(1,1)$ is defined in the positive quadrant; it cannot be continued through $xy=0$ using this formula. Taking $a,b>0$ keeps both endpoints in that component. Their ratios $y/x$ are both $b/a$, so the $G$ term cancels and
$$
\boxed{\phi(ab,b^2)-\phi(a,b)=\int_{ab}^{ab^3}s^{-2}\,ds
=\frac{b^2-1}{ab^3}.}
$$
No condition on $g$ beyond existence on the traversed ratios enters this difference.
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