Solving an exact differential equation (using an integrating factor) Given this differential equation
$x^2y^3+y+(x^3y^2-x)y'=0$
I have to find an integrating factor, such that the equation becomes an exact differential equation.
I am pretty sure that the integrating factor must be $\mu (xy)$, so let $xy=z$.
After some algebra I got this differential equation:
$\mu '(z)(z+1)(y-x)=-2\mu (z)$
I have checked my calculations many times and I am certain, that this equation is correct.
However, I don't know what to do now, since I have my $\mu$ depending on the product $xy$.
How do I get $\mu (z)$? Or am I going in the wrong direction with this attempt?
 A: 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). 
