Discuss how the function M(x, y) can be found so that each differential equation is exact:
$$M(x,y)dx+(x{e^x}^y+2xy+\frac{1}{x})dy=0$$
I understand well how the method to solve exact differential equations works, but I can not understand this problem.
Remember a proposition that states that a DE $$M(x,y)dx+N(x,y)dy=0$$ under certain hypothesis about $M(x,y)$ and $N(x,y)$, is exact iff $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$ Since you know that $N(x,y)=xe^{xy}+2xy+\frac1x$ it is needed that $$\frac{\partial M}{\partial y}=e^{xy}+xye^{xy}+2y-x^{-2}$$