Please forgive me for asking such a fundamental question.
I know for an exact differential $f(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$, we have $$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}.$$
Now if a differential $U(x,y)dx+V(x,y)dy$ is exact, then we certainly have $$\frac{\partial U}{\partial y}=\frac{\partial V}{\partial x}. \tag{1}$$
My question is, how to use Eq.(1) to prove that the differential $U(x,y)dx+V(x,y)dy$ is exact?