I am given the differential equation
$$4x^2y + 3xy^2 + 2y^3 + (2x^3 + 3x^2y + 4xy^2)\frac{dy}{dx} = 0$$
and I am asked to show that this equation has an integrating factor $\mu \equiv \mu(xy)$. Setting $M(x,y) = 4x^2y + 3xy^2 + 2y^3$ and $N(x,y) = 2x^3 + 3x^2y + 4xy^2$ I know that I want to show that, for some $\mu$ that is a function of $xy$, I have that
$$\mu(xy) M(x,y) + \mu(xy) N(x,y)\frac{dy}{dx} = 0$$ is an exact equation, i.e.
$$\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$$
setting $v = xy$ and taking the derivatives, knowing that
$$\frac{\partial \mu}{\partial x} = y\frac{\partial \mu}{\partial v}, \frac{\partial \mu}{\partial y} = x\frac{\partial \mu}{\partial v}$$
I will eventually get to
$$\frac{\mu '}{\mu} = \frac1{xy}$$
where I wrote $\mu '$ instead of $\frac{\partial \mu}{\partial v}$.
My question is, is that enough to show that the equation indeed has an integrating factor $\mu(xy)$? Or do I need to explicitly solve for $\mu$ and show that it depends only on $(xy)$?