So, you can try multiplying both sides of the equation by $1/(xy)$.
Overall, your method using $\mu(z)$ (for $z=xy$) is correct, but you get a strange result $\mu'(z)(z+1)(y-x)=-2\mu(z)$, and I don't know how you got that. I get a very different answer when using your method. It is indeed a simple ODE for $\mu(z)$, and indeed leads to the same result as just directly multiplying by $1/(xy)$.
I got my answer of multiplying by $1/(xy)$ just by playing around. But you can do it more systematically your way:
Theory: Suppose $a(x,y) + b(x,y)y' = 0$. Multiply by $\lambda(x,y)$ to get: $$ \lambda(x,y)a(x,y) + \lambda(x,y)b(x,y)y'=0$$ We want a function $f(x,y)$ such that: \begin{align} &\partial f/\partial x = \lambda(x,y)a(x,y)\\ &\partial f/\partial y = \lambda(x,y)b(x,y) \end{align} If we find such, we can say $f(x,y(x))=C$. We should have $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$. So we want: $$ \frac{\partial [\lambda(x,y)a(x,y)]}{\partial y} = \frac{\partial [\lambda(x,y)b(x,y)]}{\partial x} $$ In this case you are told to try $\lambda(x,y)=\mu(z)$ for $z=xy$ (and $\partial z/\partial x$ and $\partial z/\partial y$ can easily be calculated) and so: $$ \frac{\partial [\mu(z)(x^2y^3 +y)]}{\partial y} = \frac{\partial [\mu(z)(x^3y^2-x)]}{\partial x} $$ Doing the derivatives and simplifying leads to a simple ODE in $\mu(z)$ (with no $x$ or $y$ variables).
