$y'[x] = \frac{y}{x-y}$ Non-linear differential equation or not? $\displaystyle \qquad y'[x] = \frac{y}{x-y}=\frac{ \frac{y}{x} }{ 1-\frac{y}{x} }=\frac{u}{1-u}$
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?
 A: As Shai pointed out in the comments, the factor $x$ is missing in your update. It should be
$$xu'=\frac{u^2}{1-u}\,$$
and thus
$$
\begin{eqnarray}
u'\frac{1-u}{u^2}
&=&
\frac{1}{x}\;,
\\
-\frac{1}{u}-\log u
&=&
\log x + c\;,
\end{eqnarray}
$$
which leads to Shai's result if you plug in $u=y/x$ and cancel $\log x$. To get to the WolframAlpha result, you can switch to $v=-1/u=-x/y$ to get
$$
\begin{eqnarray}
v+\log(-v)
&=&
\log x + c\;,
\\
v\mathrm e^v
&=&
-\mathrm e^c x\;,
\\
v
&=&
W(-\mathrm e^c x)\;,
\\
-\frac{x}{y}
&=&
W(-\mathrm e^c x)\;,
\\
y
&=&
-\frac{x}{W(-\mathrm e^c x)}\;.
\end{eqnarray}
$$
Note that there's a typo in your rendition of WolframAlpha's result; the $x$ shouldn't be in the exponent.
A: You may prefer the following approach. 
First rewrite the equation as $ - y'y = y - y'x$, leading to
$$
 - \frac{{y'}}{y} = \frac{{y - y'x}}{{y^2 }},
$$
and in turn
$$
 - \frac{d}{{dx}}\log y = \frac{d}{{dx}}\frac{x}{y}.
$$
Integrating both sides then gives
$$
 - \log y = \frac{x}{y} + C.
$$
