<s>[INST] If $y=x\log\left(\frac{x}{a+bx}\right)$, then prove that $x^3\frac{d^2y}{dx^2}=\left(x\frac{dy}{dx}-y\right)$ 
If $y=x\log\left(\dfrac{x}{a+bx}\right)$, then prove that $x^3\dfrac{d^2y}{dx^2}=\left(x\dfrac{dy}{dx}-y\right)$.

My attempt: 
By differentiating, I get  
$$x\frac{dy}{dx}-y=\frac{a}{a+bx}$$ 
On differentiating again, I got  
$$x^3\frac{d^2y}{dx^2}=\frac{-abx^2}{(a+bx)^2}$$ 
Clearly, I somehow cannot equalise the LHS and RHS of the equation. Please help.
 [/INST] If $$y=x \log \left(\frac{x}{a+b x}\right)$$ $$\frac{dy}{dx}=\frac{a}{a+b x}+\log \left(\frac{x}{a+b x}\right)$$
$$\frac{d^2y}{dx^2}=\frac{a^2}{x (a+b x)^2}$$
$$x^3\frac{d^2y}{dx^2}=\frac{a^2 x^2}{ (a+b x)^2}$$
$$x\dfrac{dy}{dx}-y=\frac{a x}{a+b x}$$ So it seems that there is a typo somewhere since we find that
$$x^3\frac{d^2y}{dx^2}=\left(x\dfrac{dy}{dx}-y\right)^\color{red}{\Large2}$$
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