We have,
$$
y=\left(\sin ^{-1} 2 x\right)^2+\left(\cos ^{-1} 2 x\right)^2
$$
On differentiating both sides w.r.t. $x$, we get
$$
\frac{d y}{d x}=2 \sin ^{-1} 2 x \cdot \frac{1}{\sqrt{1-(2 x)^2}} \cdot 2+2 \cos ^{-1} 2 x
$$
$$
\left(\frac{-1}{\sqrt{1-(2 x)^2}}\right) \cdot 2
$$

$$
\Rightarrow \sqrt{1-4 x^2} \frac{d y}{d x}=4\left(\sin ^{-1} 2 x-\cos ^{-1} 2 x\right)
$$
Again differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \sqrt{1-4 x^2} y_2+\frac{1}{2 \sqrt{1-4 x^2}}(-8 x) y_1 \\
& =4\left(\frac{1}{\sqrt{1-4 x^2}} \cdot 2+\frac{1}{\sqrt{1-4 x^2}} \cdot 2\right) \\
& =\sqrt{1-4 x^2} y_2-\frac{4 x}{\sqrt{1-4 x^2}} y_1=8 \cdot \frac{2}{\sqrt{1-4 x^2}} \\
\Rightarrow \quad & \quad\left(1-4 x^2\right) y_2-4 x y_1=16
\end{aligned}
$$