The domain of the function $f\left(x\right)=\frac{1}{\sqrt{x^{12}-x^9+x^4-x+1}}$ is given by

$f\left(x\right)=\frac{1}{\sqrt{x^{12}-x^9+x^4-x+1}}$

$f\left(x\right)$ is defined for $x^{12}-x^9+x^4-x+1>0$

$\begin{array}{r}\Rightarrow x^4\left(x^8+1\right)-x\left(x^8+1\right)+1>0\\ \Rightarrow \left(x^8+1\right)x\left(x^3-1\right)+1>0\end{array}$

If $x\ge 1\text{ or }x\le -1$ then the above expression is positive.

If $-1<x\le 0,$ the above inequality still holds 

If $0<x<1,$ then $x^{12}-x\left(x^8+1\right)+\left(x^4+1\right)>0$

$\left[\because x^4+1>x^8+1\text{ and so }{x}^4+1>x\left(x^8+1\right)\right]$

The domain of $f=(-\mathrm{\infty },\mathrm{\infty })$