$$
\text { Given, } y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)
$$
Put $a x=\tan \theta$
$$
\begin{aligned}
\therefore \quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\
& =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \\
& =\tan ^{-1}\left(\tan ^{\frac{\theta}{2}}\right) \\
& =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} a x \\
\therefore \quad y & =\frac{1}{2} \tan ^{-1} a x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
y^{\prime}=\frac{1}{2\left(1+a^2 x^2\right)}
$$
Again, differentiating both side we get
$$
\begin{aligned}
y^{\prime \prime} & =-\frac{1}{2} \frac{\left(a^2 2 x\right)}{\left(1+a^2 x^2\right)^2} \\
\Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime} & =-\frac{a^2 x}{1+a^2 x^2} \\
& =-a^2 x\left(2 y^{\prime}\right) \quad[\because \text { from Eq. (i) }] \\
\Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime} & +2 a^2 x y^{\prime}=0
\end{aligned}
$$