If $y=\log {\left(\frac{x}{a+bx}\right)}^x\text{ then }{x}^3\frac{d^{2}y}{d{x}^2}=$ 

$y=x(\log x-\log (a+bx\left)\right)\text{ and differentiate }$

$\frac{\left(x\frac{dy}{dx}\right)-y}{x^2}=\frac{1}{x}-\frac{b}{a+bx}=\frac{a}{x(a+bx)}$

$\therefore x\frac{dy}{dx}-y=\frac{ax}{a+bx}$

$\text{ Differentiating again w.r.t. }\mathit{x}\text{, we get }$

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-\frac{dy}{dx}=\frac{(a+bx)a-axb}{(a+bx{)}^2}$

$\text{ or }x\frac{d^{2}y}{d{x}^2}=\frac{a^2}{(a+bx{)}^2}$

$\mathrm{or} x^3\frac{d^{2}y}{d{x}^2}=\frac{a^2{x}^2}{(a+bx{)}^2}={\left(x\frac{dy}{dx}-y\right)}^2$