Explanation

(a)
Given, $\frac{dy}{dx} = - \frac{3xy + y^{2}}{x^{2} + xy}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\Rightarrow$ $v + x\ \ \frac{dv}{dx} = - \frac{3v + v^{2}}{1 + v}$
$\Rightarrow$ $x\ \frac{dv}{dx} = \frac{- 2v(v + 2)}{v + 1}$
$\Rightarrow$ $\frac{1}{x}\ dx = - \frac{(v + 1)}{2v(v + 2)}\ dv$
$\Rightarrow$ $- \frac{2}{x}\ dx = - \left\lbrack \frac{1}{2(v + 2)} + \frac{1}{2v} \right\rbrack dv$
$\Rightarrow$ $- 2\log_{e}x = \frac{1}{2}\log{(v + 2)} + \frac{1}{2}\log v - \log c$
$\Rightarrow$ $v(v + 2)x^{4} = c^{2}$
$\Rightarrow$ $\frac{y}{x}\left( \frac{y}{x} + 2 \right)x^{4} = c^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \left( y^{2} + 2xy \right)x^{2} = c^{3}$