Given $y^{2}dx+\left(x^2-xy+y^2\right)dy=0$

$\Rightarrow \frac{dx}{dy}=-\frac{\left(x^2-xy+y^2\right)}{y^2} ...\left(\mathrm{i}\right)$

Now let $x=vy$  $\Rightarrow v+y\frac{dv}{dy}=\frac{dx}{dy}$

Now putting in equation $\left(\mathrm{i}\right)$ we get

$\Rightarrow v+y\frac{dv}{dy}=-\left(v^2-v+1\right)$

$\Rightarrow y\frac{dv}{dy}=-v^2-1$

$\Rightarrow \frac{-dv}{1+v^2}=\frac{dy}{y}$ 

Now Integrating Both side

$\Rightarrow -\int \frac{dy}{1+v^2}=\int \frac{dy}{y}$

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

Also  this curve passes through $\left(1,1\right)$ 

$\Rightarrow {\tan }^{-1}1=\log \left|1\right|+c$ $\Rightarrow c=-\frac{\pi }{4}$

$\Rightarrow -{\tan }^{-1}\frac{x}{y}=\log \left|y\right|-\frac{\pi }{4}$

Now putting $y=\sqrt{3}x$ and $x=\alpha$ we get

$=-{\tan }^{-1}\frac{x}{\sqrt{3}x}=\log \left|\sqrt{3}\alpha \right|-\frac{\pi }{4}$ $=-{\tan }^{-1}\frac{1}{\sqrt{3}}=\log \left(\sqrt{3}\alpha \right)-\frac{\pi }{4}$

$=-\frac{\pi }{6}+\frac{\pi }{4}=\log \left|\sqrt{3}\alpha \right|$

  $\Rightarrow \log \left|\sqrt{3}\alpha \right|=\frac{\pi }{12}$