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The given differential equation is:

$y \log y d x-x d y=0$

$\Rightarrow y \log y d x=x d y$

$\Rightarrow \frac{d y}{y \log y}=\frac{d x}{x}$

Integrating both sides, we get:

$\int \frac{d y}{y \log y}=\int \frac{d x}{x}$              ...(1)

Let $\log y=t$.

$\therefore \frac{d}{d y}(\log y)=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y}=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y} d y=d t$

Substituting this value in equation (1), we get:

$\int \frac{d t}{t}=\int \frac{d x}{x}$

$\Rightarrow \log t=\log x+\log \mathrm{C}$

$\Rightarrow \log (\log y)=\log \mathrm{C} x$

$\Rightarrow \log y=\mathrm{C} x$

$\Rightarrow y=e^{\mathrm{Cx}}$

This is the required general solution of the given differential equation.