Given $y= (\sin ^{-1}x)^2+A\cos ^{-1}x+B$
Let us first differentiate with respect to x on both sides
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}$$+A\bigg(\large\frac{-1}{\sqrt {1-x^2}}\bigg)+0$
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}-\frac{A}{\sqrt {1-x^2}}$
This can be written as
$\sqrt {1-x^2}.\large\frac{dy}{dx}$$=2 \sin ^{-1}x-A$
Now again differentiating on both sides w.r.t x
Since $\sqrt {1-x^2}.\large\frac{dy}{dx}$ are two functions in the product form, we can apply the product rule $\large\frac{d}{dx}$$(uv)=u.\large\frac{d}{dx}$$(v)+v.\large\frac{d}{dx}$$(u)$
Here let $u=\sqrt {1-x^2},$ hence $ \large\frac{d}{dx}$$(v)=\large\frac{1}{2 \sqrt {1-x^2}}$$(-2x)$
$v=\large\frac{dy}{dx}\qquad \frac{d}{dx}$$(v)=\large\frac{d^2y}{dx^2}$
$\large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}+\frac{dy}{dx}.\frac{-2x}{2\sqrt {1-x^2}}$$=2 \large\frac{1}{\sqrt {1-x^2}}-0$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}-\frac{x}{\sqrt {1-x^2}}.\frac{dy}{dx}$$= \large\frac{2}{\sqrt {1-x^2}}$
$\Large\frac{(1-x^2)\bigg(\frac{d^2y}{dx^2}\bigg)-x\bigg(\frac{dy}{dx}\bigg)}{\sqrt {1-x^2}}=\frac{2}{\sqrt {1-x^2}}$
=>$(1-x^2)\large\frac{d^2y}{dx^2}$$-x \large\frac{dy}{dx}$$-2=0$