Explanation

(b)
Given, $\frac{y}{x} = \log{x - \log{(a + bx)}}$
$\Rightarrow \frac{x\frac{dy}{dx} - y}{x^{2}} = \frac{1}{x} - \frac{1}{a + bx}b = \frac{a}{x(a + bx)}\$
$\Rightarrow \ \ x\frac{dy}{dx} - y = \frac{ax}{a + bx}\ \ \ \ \$…(i)
$\Rightarrow \ x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{(a + bx)a - ax.b}{(a + bx)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
$\Rightarrow x\frac{d^{2}y}{dx^{2}} = \frac{a^{2}}{(a + bx)^{2}}$
$\Rightarrow \ x^{3}\frac{d^{2}y}{dx^{2}} = \left( x\frac{dy}{dx} - y \right)^{2}\$ [using eq(i)]