How can i solve a differential equation like this one? My Problem is: this given differential equation
$$x^3+y^3+x^2y-xy^2y^{\prime}=0$$ $$(x\neq 0,\ y\neq 0)$$ 
My Approach was: i had the idea to bring it in this form:
$$x^3+y^3+x^2y-xy^2y^{\prime}=0$$
$$x^3+y^3+x^2y=xy^2y^{\prime}$$
$$\frac{x^3}{xy^2}+\frac{y^3}{xy^2}+\frac{x^2y}{xy^2}=\frac{xy^2y^{\prime}}{xy^2}$$
$$\frac{x^3}{xy^2}+\frac{y^3}{xy^2}+\frac{x^2y}{xy^2}=y^{\prime}$$
$$\frac{x^2}{y^2}+\frac{y}{x}+\frac{x}{y}=y^{\prime}$$
But this is the point where i get stuck. It seems the expression is gettin more and more complex, and is not leading to any solution... how can i solve this?
 A: Hint: Make the change of variable $y=zx$. You will get a separable equation. You can do that from the point you reached, but it is quite a bit simpler to start all over again. For general information (which you will not need in this case) search for homogeneous differential equation.
More: Let $y=zx$. Then $y'=z+xz'$. Substituting in the equation, we get 
$$x^3+z^3x^3 +x^3z-x^3 z^2(z+xz')=0.$$
Divide through by $x^3$. There is some nice cancellation, and we end up with
$$xz^2z'=1+z.$$
This is a separable equation, which I expect you can handle. There is a complication in that we will end up with an implicit solution. 
A: Since you have expressed y' in terms of a function of $\frac{y}{x}$, you can do a substitution: set v = $y\over x$, and $\frac{dy}{dx} = v + x\frac{dv}{dx}$, and you get $$v + x\frac{dv}{dx} = v^{-2} + v^{-1} + v \longrightarrow x\frac{dv}{dx} = v^{-2} + v^{-1}.$$Separate the equation:$$\frac{dv}{v^{-2} + v^{-1}} = \frac{dx}{x}$$
Integrate both sides:
$$\ln\left(v+1\right)+\frac{\left(v-2\right)v}{2}+C = \ln x$$
Now solve for v, undo the substitution, and solve for y.
