Given, differential equation is
$ y(1+\log x) \frac{d x}{d y}-x \log x=0$
$ \Rightarrow \frac{(1+\log x) d x}{x \log x}=\frac{d y}{y} $
$ \Rightarrow \left(\frac{1}{x \log x}+\frac{1}{x}\right) d x=\frac{1}{y} d y$
On integrating both sides, we get
$\int\left(\frac{1}{x \log x}+\frac{1}{x}\right) d x=\int \frac{1}{y} d y$
Put $\log x=t$
$\Rightarrow \frac{1}{x} d x=d t$
$\therefore \int \frac{1}{t} dt+\int \frac{1}{x} d x=\int \frac{1}{y} dy$
$\Rightarrow \log t+\log x=\log y+\log c$
$\Rightarrow \log t x=\log y c$
$\Rightarrow t x=y c$
$\Rightarrow x \log x=y c$
When $x=e$ and $y=e^{2}$
$\therefore e \log e=e^{2} c$
$\Rightarrow e \times 1=e^{2} c$
$\Rightarrow c \Rightarrow \frac{1}{e}$
$\therefore x \log x=\frac{y}{e}$
$\Rightarrow y=e x \log x$