If ${\log }_{1/3}\left({\log }_4\left({\mathrm{x}}^2-5\right)>0\right.$, then 

$\begin{array}{l}\text{ base }=\frac{1}{3},\log \text{ is decreasing }\Rightarrow {\log }_4^{\left({\mathrm{x}}^2-5\right)}<1\\ \Rightarrow {\mathrm{x}}^2-5<4{\mathrm{x}}^2-9<0\end{array}$
$\Rightarrow \mathrm{x}\in (-3,3) \mathrm{but} \mathrm{for} \mathrm{domain} {\log }_4^{\left({\mathrm{x}}^2-5\right)}>0$
i.e.$,{\mathrm{x}}^2-5>1\Rightarrow \mathrm{x}\in (-\mathrm{\infty },-\sqrt{6})\cup (\sqrt{6},\mathrm{\infty })$