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$y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

$\Rightarrow y d x=\left[2 x-x \log \left(\frac{y}{x}\right)\right] d y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}$             ...(1)

Let $F(x, y)=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log \left(\frac{\lambda y}{\lambda x}\right)}=\frac{y}{2 x-\log \left(\frac{y}{x}\right)}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log v}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v}$

$\Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}$

$\Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}$

$\Rightarrow\left[\frac{1+(1-\log v)}{v(\log v-1)}\right] d v=\frac{d x}{x}$

$\Rightarrow\left[\frac{1}{v(\log v-1)}-\frac{1}{v}\right] d v=\frac{d x}{x}$

Integrating both sides, we get:

$\int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x$

$\Rightarrow \int \frac{d v}{v(\log v-1)}-\log v=\log x+\log \mathrm{C}$        $\ldots(2)$

$\Rightarrow$ Let $\log v-1=t$

$\Rightarrow \frac{d}{d v}(\log v-1)=\frac{d t}{d v}$

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

$\Rightarrow \frac{d v}{v}=d t$

Therefore, equation (1) becomes:

$\Rightarrow \int \frac{d t}{t}-\log v=\log x+\log \mathrm{C}$

$\Rightarrow \log t-\log \left(\frac{y}{x}\right)=\log (\mathrm{C} x)$

$\Rightarrow \log \left[\log \left(\frac{y}{x}\right)-1\right]-\log \left(\frac{y}{x}\right)=\log (\mathrm{C} x)$

$\Rightarrow \log \left[\frac{\log \left(\frac{y}{x}\right)-1}{\frac{y}{x}}\right]=\log (\mathrm{C} x)$

$\Rightarrow \frac{x}{y}\left[\log \left(\frac{y}{x}\right)-1\right]=\mathrm{C} x$

$\Rightarrow \log \left(\frac{y}{x}\right)-1=\mathrm{C} y$

This is the required solution of the given differential equation.