If the solve the problem

Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$

Given,

$y=\left(\sin ^{-1} x\right)^{2} \ldots \ldots$ equation 1

to prove : $\left(1-x^{2}\right) y_{2}-x y_{1}-2=0$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

$\mathrm{As}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $\mathrm{dy} / \mathrm{dx}$

$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x\right)^{2}$

Using chain rule we will differentiate the above expression

Let $t=\sin ^{-1} x=>\frac{d t}{d x}=\frac{1}{\sqrt{\left(1-x^{2}\right)}}$ [using formula for derivative of $\sin ^{-1} x$ ]

And $y=t^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{d y}{d x}=2 t \frac{1}{\sqrt{\left(1-x^{2}\right)}}=2 \sin ^{-1} x \frac{1}{\sqrt{\left(1-x^{2}\right)}} \ldots \ldots . .$ equation 2

Again differentiating with respect to $x$ applying product rule:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \sin ^{-1} \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)+\frac{2}{\sqrt{\left(1-\mathrm{x}^{2}\right)}} \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}$

$\frac{d^{2} y}{d x^{2}}=-\frac{2 \sin ^{-1} x}{2\left(1-x^{2}\right) \sqrt{1-x^{2}}}(-2 x)+\frac{2}{\left(1-x^{2}\right)}\left[\right.$ using $\left.\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{\left(1-x^{2}\right)}}\right]$

$\frac{d^{2} y}{d x^{2}}=\frac{2 x \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{2}{\left(1-x^{2}\right)}$

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=2+\frac{2 x \sin ^{-1} x}{\sqrt{1-x^{2}}}$

Using equation 2 :

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=2+x \frac{d y}{d x}$

$\therefore\left(1-x^{2}\right) y_{2}-x y_{1}-2=0 \ldots \ldots$ proved