Given, $\left(a+bx\right) e^{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=x$

$\Rightarrow e^{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=\frac{x}{\left(a+bx\right)}$

$\Rightarrow \log \left(e^{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)=\log \left(\frac{x}{a+bx}\right)$

$\Rightarrow \frac{y}{x}=\log x-\log \left(a+bx\right)$

$\Rightarrow y=x\log x-x\log \left(a+bx\right)$

Differentiate both sides

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

$y\text{'}=x\left[\frac{1}{x}-\frac{b}{a+bx}\right]+\left[\log x-\log \left(a+bx\right)\right]$

$y\text{'}=x\left[\frac{1}{x}-\frac{b}{a+bx}\right]+\frac{y}{x}$

$xy\text{'}=x^2\left[\frac{1}{x}-\frac{b}{a+bx}\right]+y$

$xy\text{'}=x^2\left[\frac{a+bx-bx}{x\left(a+bx\right)}\right]+y$

$xy\text{'}=\left[\frac{ax}{\left(a+bx\right)}\right]+y$

$x\frac{d^{2}y}{d{x}^2}+y\text{'}=a\left[\frac{\left(a+bx\right)·1-x·b}{{\left(a+bx\right)}^2}\right]+y\text{'}$

$x^3\frac{d^{2}y}{d{x}^2}=\left[\frac{a^2{x}^2}{{\left(a+bx\right)}^2}\right]$

$x^3\frac{d^{2}y}{d{x}^2}={\left[\frac{ax}{\left(a+bx\right)}\right]}^2$

$x^3\frac{d^{2}y}{d{x}^2}={\left[xy\text{'}-y\right]}^2$

$\frac{d^{2}y}{d{x}^2}=\frac{1}{x^3}{\left[xy\text{'}-y\right]}^2$