$y\log ydx - xdy = 0$

.: We have, $y\log ydx - xdy = 0$
$\Rightarrow y\log ydx = xdy \Rightarrow \cfrac{{dx}}{x} = \cfrac{{dy}}{{y\log y}}$

…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{dx}}{x}} = \int {\cfrac{{dy}}{{y\log y}}}$
$\Rightarrow \log x = \int {\cfrac{{dz}}{z}}$

[Let $\log y = z \Rightarrow \cfrac{1}{y}dy = dz$]
$\Rightarrow \log x + \log C = \log z$

$\Rightarrow \log (Cx) = \log z \Rightarrow Cx = z \Rightarrow Cx = \log y \Rightarrow y = {e^{Cx}}$

which is the required solution.