How do I solve $y'=\frac{y}{x}\frac{x-y}{x+y}$? I have a solution for  
$y'=\frac{x-y}{x+y}$
Now I have this new problem:
$y'=\frac{y}{x}\frac{x-y}{x+y}$
My first approach was to solve it by parts; that way I could have re-used my solution for $\frac{x-y}{x+y}$.  However, that solution is implicit in $y$, so I couldn't re-use it.
Then I tried making the problem homogeneous in $\frac{y}{x}$, but I ended up with a loose $y$ term.
Now I'm out of ideas.  Can someone show me how it's done?

Let $v=y/x$ or $y=vx$. Then by the product rule $y'=v+v'x$ and by substitution
$$v+v'x = v \frac{1+v}{1-v}$$
$$v'x=\frac{v+v^2}{1-v}-\frac{v-v^2}{1-v}=2\frac{v^2}{1-v}$$
and it remains to integrate.
