The general solution of $y^2 d x+\left(x^2-x y+y^2\right) d y=0$, is

$\log y=\tan ^{-1} \frac{y}{x}+C$

We have,

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

Putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$, we get

$v+x \frac{d v}{d x}=-\frac{v^2}{1-v+v^2}$

$\Rightarrow x \frac{d v}{d x}=\frac{-v-v^3}{v^2-v+1}$

$\Rightarrow \frac{v^2-v+1}{v\left(v^2+1\right)} d v=-\frac{d x}{x}$

$\Rightarrow \left(\frac{1}{v}-\frac{1}{v^2+1}\right) d v=-\frac{d x}{x}$

On integrating, we get

$\log v-\tan ^{-1} v=-\log x+C$

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