Solution
= Solution
The two coefficients are the corresponding <partial derivatives> of $e^{x+y}$. Therefore the <differential form> is exact, with
$$
\boxed{f(x,y)=e^{x+y}+C,\qquad df=e^{x+y}dx+e^{x+y}dy.}
$$
The arbitrary constant does not change the <total differential>.