If $|z|=1,|a|\ne 1$, and $x=\frac{z}{(z-a)(1-z\bar{a})}$, then

Given $\left|z\right|=1$  $,$  $\left|a\right|\ne 1$

then, $x=\frac{z}{\left(z-a\right)\left(1-z\bar{a}\right)}=\frac{1}{\left(1-a\right)\left(1-a\right)}$

let $a=x+iy$  and  $\bar{a}=x-iy$

$=\frac{1}{\left[1-\left(x+iy\right)\right]\left[1-\left(x-iy\right)\right]}$

$=\frac{1}{1-\left(x-iy\right)-\left(x+iy\right)+\left(x+iy\right)\left(x-iy\right)}$

$=\frac{1}{1-x+iy-x-iy+x^2-{\left(iy\right)}^2}$

$=\frac{1}{{\left(x-1\right)}^2-y^2}$

$\therefore x>0.$

Hence, the answer is $x>0.$