The domain of the function given by  $f\left(x\right)={\log }_3\left({\log }_{\frac{1}{2}}\left({\log }_2\left({\log }_{\frac{1}{3}}\left(x\right)\right)\right)\right)+\frac{x+1}{\sqrt{1-3x-4x^2}}is$

$f\left(x\right)={\log }_3\left({\log }_{1/2}\left({\log }_2\left({\log }_{1/3}\right)\right)\right)$

$\frac{x+1}{\sqrt{1-3x-4x^2}}$

$Now,\sqrt{1-3x-4x^2}>0$

$4x^2+3x-1<0$

$\therefore (4x-1)(x+1)<0$

${\log }_{1/3}\left(x\right)<2$

$x>\frac{1}{9}$

$nowF\left(f\left(x\right)\right)\in x>\frac{1}{9}\cap x\in (-1,1/4)$

$\therefore x\in (\frac{1}{9},\frac{1}{4})$