Explanation

(c)
$\because y = \left( \frac{ax + b}{cx + d} \right)$
or $c\ xy + dy = ax + b$
On differentiating both sides w.r.t. $x$, we get
$c\left\{ x\frac{dy}{dx} + y.1 \right\} + d\frac{dy}{dx} = a$
or $x\frac{dy}{dx} + y + \left( \frac{d}{c} \right)\frac{dy}{dx} = \left( \frac{a}{c} \right)$
Again differentiating both sides w.r.t. $x$, we get
$x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} + \frac{dy}{dx} + \left( \frac{d}{c} \right)\frac{d^{2}y}{dx^{2}} = 0$
or $x + \frac{2\frac{dy}{dx}}{\left( \frac{d^{2}y}{dx^{2}} \right)} + \frac{d}{c} = 0$
Again, on differentiating both sides w.r.t. $x$, we get
$1 + \frac{\left( \frac{d^{2}y}{dx^{2}}.2\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx}.\frac{d^{3}y}{dx^{3}} \right)}{\left( \frac{d^{2}y}{dx^{2}} \right)} + 0 = 0$
$\Rightarrow 2\frac{dy}{dx}.\frac{d^{3}y}{dx^{3}} = 3\left( \frac{d^{2}y}{dx^{2}} \right)^{2}$