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I am trying to find the general solution to the following equation, but the integral at the end is very complicated and leads me to believe I may have made a mistake somewhere.

$$xy^2\frac {dy}{dx} = y^3 + xy^2 -x^2y - x^3 $$

Which, by the substitution $z = \frac {y}{x}$, can be rearranged into the equation

$$x\frac {dz}{dx} = 1 - \frac {1}{z} - \frac {1}{z^2} $$

This is a separable equation, which I separated into

$$ \int \frac {1}{1 - \frac {1}{z} - \frac {1}{z^2}}dz = \int \frac {1}{x}dx $$

The right-hand side is easy to solve, but the left-hand integral is giving me trouble. Assuming I did the steps leading up to it correctly, the integral has me stumped. Even WolframaAlpha is unhelpful. My first thought would be to try a partial fraction, but after a few attempts it does not seem to work.

Am I approaching this differential equation correctly? Is there an error I haven't caught?

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    @Moo that is what I did. Substituting that, then subtracted $z$ from both sides, gives the second equation. – 2017-02-08

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Well, assuming that this is the integral to solve, use that $$\frac{1}{ 1-\frac{1}{z}-\frac{1}{z^2}}= \frac{z^2}{z^2-z-1}=1+\frac{z+1}{z^2-z-1}. $$ That $1$ is good as gone. For the remaining part:

$$ \frac{z+1}{z^2-z-1}=\frac{1}{2}\frac{2z+2}{z^2-z-1} = \frac{1}{2}\left (\frac{2z-1}{z^2-z-1} + \frac{3}{z^2-z-1} \right),$$ and substituition solves the first fraction. For the second, the usual approach using partial fractions will do the trick, but the roots of $z^2-z-1$ seem ugly. Just don't panic.