The term independent of x in the expansion of ${\left[(t^{-1}-1)x+{(t^{-1}+1)}^{-1}{x}^{-1}\right]}^8$  is 

${\left[\left(t^{-1}-1\right)x+{\left(t^{-1}+1\right)}^{-1}{x}^{-1}\right]}^8$

$={\left[\left(\frac{{\displaystyle 1}}{{\displaystyle t}}-1\right)x+{\left(\frac{{\displaystyle t+1}}{{\displaystyle t}}\right)}^{-1}\frac{{\displaystyle 1}}{{\displaystyle x}}\right]}^8$

$={\left[\left(\frac{1-t}{t}\right)x+\left(\frac{t}{t+1}\right)\frac{1}{x}\right]}^8$

$T_{r+1}={}^{\text{8}}{\text{C}}_r{\left[\left(\frac{{\displaystyle 1-t}}{{\displaystyle t}}\right)\text{ x }\right]}^{8-r}\left[{\left(\frac{{\displaystyle t}}{{\displaystyle t+1}}\right)\frac{1}{x}}^r\right]$

$={}^{\text{8}}{\text{C}}_r\frac{{(1-t)}^{8-r}}{t^r}.\frac{t^r}{{(t+1)}^r}.\frac{x^{8-r}}{x^r}$

$={}^{\text{8}}{\text{C}}_r{(1-t)}^{8-r}.{(1+t)}^{-r}.x^{8-2r}$

For independent term

$8-2r=0$

$r=4$

$\therefore$ Independent term$={}^{\text{8}}{\text{C}}_4{(1-t)}^{8-4}.{(1+t)}^{-4}$

$=\frac{8\times 7\times 6\times 5}{1\times 2\times 3\times 4}$

$=70{\left(\frac{{\displaystyle 1-t}}{{\displaystyle 1+t}}\right)}^4$