$Domainoff\left(x\right)=\sqrt{e^{\sin {}^{-1}\left(\log {}_{16}{x}^2\right)}}is$

$\begin{array}{l}f\left(x\right)=\sqrt{e^{{\sin }^{-1}\left(\log {}_{16}{x}^2\right)}}isdefined\\ iff-1\le {\log }_{16}{x}^2\le 1\\ \Rightarrow {16}^{-1}\le x^2\le 16\\ \Rightarrow \frac{1}{16}\le x^2\le 16\\ Let{x}^2\le 16\Rightarrow (x+4)(x-4)\le 0\\ \therefore x\in [-4,4]\\ and{x}^2\ge \frac{1}{16}\Rightarrow (x+\frac{1}{4})(x-\frac{1}{4})\ge 0\\ x\in (-\infty ,\frac{1}{4}]\cup [\frac{1}{4},\infty )\\ \end{array}$

$\therefore x\in [-4,\frac{1}{4}]\cup [\frac{1}{4},4]$