Explanation

(a)
Given, $\frac{dy}{dx} = \frac{- y^{2}}{x^{2} - xy + y^{2}}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \ \frac{dv}{dx}$
$\therefore$ $v + x\ \ \ \frac{dv}{dx} = \frac{- v^{2}}{v^{2} - v + 1}$
$x\ \ \frac{dv}{dx} = \frac{- v^{3} - v}{v^{2} - v + 1}$
$\Rightarrow$ $\frac{(v^{2} - v + 1)}{- v^{3} - v}\ dv = \frac{1}{x}\ dx$
$\Rightarrow$ $\frac{- \left( v^{2} + 1 \right) + v}{v(v^{2} + 1)}\ dv = \frac{1}{x}\ dx$
$\Rightarrow$ $\int_{}^{}{- \frac{1}{v}}dv + \int_{}^{}\frac{1}{v^{2} + 1}dv = \int_{}^{}\frac{1}{x}\ dx$
$\Rightarrow$ $- \log v + \tan^{- 1}v = \log x + c$
$\Rightarrow$ $\tan^{- 1}v = \log{xv} + c$
$\Rightarrow$ $\tan^{- 1}\left( \frac{y}{x} \right) = \log y + c$