Let $y=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)=\sin ^{-1}\left(\frac{2 \cdot 2 x}{1+(2 x)^{2}}\right)$
Put $2 x=\tan \theta \Rightarrow \theta=\tan ^{-1} 2 x$
$\therefore y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$=\sin ^{-1}(\sin 2 \theta)=2 \theta\left(\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$=2 \tan ^{-1} 2 x$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{2}{1+(2 x)^{2}} \cdot 2=\frac{4}{1+4 x^{2}}$
ALTERNATE SOLUTION
$y=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)$
$\frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{4 x}{1+4 x^{2}}\right)^{2}} \times \frac{\left(1+4 x^{2}\right) 4-4 x(8 x)}{\left(1+4 x^{2}\right)^{2}}}$
$=\frac{1+4 x^{2}}{\sqrt{\left(1+4 x^{2}\right)^{2}-16 x^{2}}} \times \frac{4+16 x^{2}-32 x^{2}}{\left(1+4 x^{2}\right)^{2}}$
$=\frac{4-16 x^{2}}{\left(1+4 x^{2}\right) \sqrt{1-8 x^{2}+16 x^{4}}}$
$=\frac{4-16 x^{2}}{\left(1+4 x^{2}\right)\left(1-4 x^{2}\right)}$
$=\frac{(2)^{2}-(4 x)^{2}}{\left(1+4 x^{2}\right)(1-2 x)(1+2 x)}=\frac{(2+4 x) 2}{\left(1+4 x^{2}\right)(1+2 x)}=\frac{4}{1+4 x^{2}}$