Solve karo bhai

Not fully solved ; but tryin' to simplify a bit...

x\{y log(x) - x\}\frac{dy}{dx} = y \{xlog(y)-y\}

y^2 - xy \ log(\frac{y}{x})= x^2 \frac{dy}{dx}

Dividing by "xy" throughout-

\frac{y}{x} - log(\frac{y}{x})= \frac{x}{y} \frac{dy}{dx}

Subs. , y/x =t
t - log(t) = \frac{1}{t} \left( x\frac{dt}{dx} +t \right)

Which is reducing our Expression to -

\frac{dx}{x} = \frac{dt}{t^2 -t-t.log(t)}

Bhai, iske aage toh bas yehi soojh rahaa hai...
log(t) = u,

So, \frac{dx}{x} = \frac{du}{(e^u -u-1)}

Toh, now reduced to an integration Qn --

Find - \int \frac{du}{(e^u -u-1)}