1. Given differential equation

$y^2\,dx+(x^2-xy+y^2)\,dy=0$

We need the particular solution passing through $(1,1)$, and then find where it meets the line

$y=\sqrt{3}\,x.$

2. Write as differential equation in a convenient form

Treat $x$ as a function of $y$:

$y^2\frac{dx}{dy}+x^2-xy+y^2=0$

So,

$\frac{dx}{dy}=-\frac{x^2-xy+y^2}{y^2}.$

This is homogeneous in $x$ and $y$. Let

$x=vy \quad \Rightarrow \quad \frac{dx}{dy}=v+y\frac{dv}{dy}.$

Substitute into the equation:

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

Hence,

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

Therefore,

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

3. Integrate

Integrating both sides,

$\tan^{-1}v=-\ln|y|+C.$

Since $v=\dfrac{x}{y}$, we get

$\tan^{-1}\left(\frac{x}{y}\right)+\ln|y|=C.$

4. Use the point $(1,1)$

At $(x,y)=(1,1)$,

$\tan^{-1}(1)+\ln 1=C.$

So,

$C=\frac{\pi}{4}.$

Thus the required solution curve is

$\tan^{-1}\left(\frac{x}{y}\right)+\ln y=\frac{\pi}{4}$

(using $y>0$ here since points involved have positive coordinates).

5. Use intersection with $y=\sqrt{3}x$

At the intersection point $(\alpha,\sqrt{3}\alpha)$,

$\frac{x}{y}=\frac{\alpha}{\sqrt{3}\alpha}=\frac{1}{\sqrt{3}}.$

Hence,

$\tan^{-1}\left(\frac{x}{y}\right)=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}.$

Also,

$y=\sqrt{3}\alpha.$

Substitute into the solution:

$\frac{\pi}{6}+\ln(\sqrt{3}\alpha)=\frac{\pi}{4}.$

Therefore,

\ln(\sqrt{3}\alpha)=\frac{\pi}{4}-\frac{\pi}{6}=rac{\pi}{12}.

6. Match with options

$\log_e(\sqrt{3}\alpha)=\frac{\pi}{12}$

So the correct option is:

C. $\dfrac{\pi}{12}$

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.