\displaystyle \qquad y'[x] = \frac{y}{x-y}=\frac{ \frac{y}{x} }{ 1-\frac{y}{x} }=\frac{u}{1-u}
so
\displaystyle \qquad u = \frac{y}{x} \rightarrow y=ux,\qquad y'[x] = u
$\displaystyle \qquad u = \frac{u}{1-u} \rightarrow u^{2}=0$
But wolframalpha states that it a first order non-linear ODE here and its solution: $y(x) = - \frac{x} {W(-e^{-C_{1} x})}$ where $W(z)$ is a product log function(?!).
Could someone explain which way is the right and which is really the solution. My friend just said that he solved it by integrating, separating $x$'s and $y$'s to different sides but I cannot get it that way.
[update] thanks to joriki: $\frac{d u}{d x} = \frac{u}{1-u}-u=\frac{u^{2}}{1-u}$ and after integrating $\frac{-1}{u}-\ln(u)-x+C=0$ but not yet the solution, thinking...but how to get to the solution suggested by WA now? I cannot see a way to solve it now just for x or just for y. Ideas?