$${{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}$$

Put $$y = vx$$ we get

$$v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}$$

$$ \Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}$$

$$ \Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2})}}dv = \int {{{dx} \over x}} } $$

$${\tan ^{ - 1}}\left( {{y \over x}} \right) - \ln \left( {{y \over x}} \right) = \ln x + c$$

As it passes through (1, 1)

$$c = {\pi \over 4}$$

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

Put $$y = \sqrt 3 x$$ we get

$$ \Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}$$

$$ \Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha $$

$$\therefore$$ $$\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha $$

$$ = \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}$$