$y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right)$
Then $\frac{d y}{d x}=x \cdot \frac{1}{\left(\frac{1}{a x}+\frac{1}{a}\right)} \cdot\left(-\frac{a}{a x^2}\right)+\log \left(\frac{1}{a x}+\frac{1}{a}\right)$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{-a x}{a x(1+x)}+\log \left(\frac{1}{a x}+\frac{1}{a}\right) \\ & \Rightarrow \frac{d y}{d x}=-\frac{1}{(1+x)}+\log \left(\frac{1}{a x}+\frac{1}{a}\right)\end{aligned}$
and $\frac{d^2 y}{d x^2}=+\frac{1}{(x+1)^2}+\frac{1}{\left(\frac{1}{a x}+\frac{1}{a}\right)} \cdot\left(-\frac{1}{a x^2}\right)$
$\begin{aligned} & =\frac{1}{(x+1)^2}-\frac{a x}{(1+x)\left(a x^2\right)} \\ & \Rightarrow \frac{d y}{d x^2}=\frac{1}{(x \quad 1)^2}-\frac{1}{x(x \quad 1)}\end{aligned}$
Now, $x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y$
$=x(x+1)\left[\frac{1}{(x+1)^2}-\frac{1}{x(x+1)}\right]$
$+x\left[-\frac{1}{(1+x)}+\log \left(\frac{1}{a x}+\frac{1}{a}\right)\right]-x \log \left(\frac{1}{a x}+\frac{1}{a}\right)$
$\begin{aligned} & =\frac{x}{x+1}-1-\frac{x}{x+1}+x \log \left(\frac{1}{a x}+\frac{1}{a}\right)-x \log \left(\frac{1}{a x}+\frac{1}{a}\right) \\ & =-1\end{aligned}$
$\therefore x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=-1$.