Explanation

(d)
${\because y = x\ In\left( \frac{x}{a + bx} \right) = x(In\ x - In(a + bx))
}{or\ \left( \frac{y}{x} \right) = In\ x - In\ (a + bx)}$
On differentiating both sides w.r.t. $x$, we get
$\left( \frac{x\frac{dy}{dx} - y.1}{x^{2}} \right) = \frac{1}{x} - \frac{b}{a + bx} = \frac{a}{x(a + bx)}\ \ \ldots(i)$
or $\left( x\frac{dy}{dx} - y \right) = \frac{ax}{a + bx}$
On taking log on both sides, we get
$In\ \left( x\frac{dy}{dx} - y \right) = In\ (ax) - In\ (a + bx)$
On differentiating both sides w.r.t. $x$, we get
$\frac{x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - \frac{dy}{dx}}{\left( x\frac{dy}{dx} - y \right)} = \frac{1}{x} - \frac{b}{a + bx} = \frac{a}{x(a + bx)}$
$= \frac{\left( x\frac{dy}{dx} - y \right)}{x^{2}}\ \ \ \lbrack from\ Eq.(i)\rbrack$
or $x^{3}\frac{d^{2}y}{dx^{2}} = \left( x\frac{dy}{dx} - y \right)^{2}$