The function $f\left(x\right)=\sqrt{x^{{\log }_{5}x}-5}$  is defined for

$f\left(x\right)=\sqrt{\left(x^{{\log }_{5}x}-5\right)}$ 

$f\left(x\right)$  is defined

if $x^{{\log }_{5}x}-5\ge 0$      and $x>0$

$\Rightarrow$   $x^{{\log }_{5}x}\ge 5$

$\Rightarrow$   ${\log }_{5}x.{\log }_{5}x\ge 1$   $\left(take \log function to the base 5 on both sides\right)$

$\Rightarrow$    $\left({\log }_{5}x-1\right)\left({\log }_{5}x+1\right)\ge 0$

$\therefore$    ${\log }_{5}x\le -1\Rightarrow x\le 5^{-1}$  or ${\log }_{5}x\ge 1\Rightarrow x\ge 5^1$

But $x>0$   

$\therefore$   $x\in (0,\frac{1}{5}]\cup [5,\infty )$