Solution
= Solution
The <chain rule> gives $d(e^{xy})=e^{xy}(y\,dx+x\,dy)$, so the given <total differential> is exact. Thus $f=e^{xy}+C$; two such <functions> differ by a constant because their <gradient> difference vanishes on the <connected> <plane>. The assigned value at the origin forces $C=-1$, giving
$$
\boxed{f(x,y)=e^{xy}-1}.
$$