Let $f\left(x\right)=ln\left(\frac{1-x}{1+x}\right)$ . The set of values of $\text{'}\alpha \text{'}$  for which $f\left(\alpha \right)+f\left({\alpha }^2\right)=f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)$  is satisfied are

$f\left(\alpha \right)+f{\left(\alpha \right)}^2=ln\left[\left(\frac{1-\alpha }{1+\alpha }\right)\left(\frac{1-{\alpha }^2}{1+{\alpha }^2}\right)\right]=ln\left[\frac{{(1-\alpha )}^2}{1+{\alpha }^2}\right]$

$f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)=ln\left[\frac{1-\frac{\alpha }{{\alpha }^2-\alpha +1}}{1+\frac{\alpha }{{\alpha }^2-\alpha +1}}\right]=ln\left[\frac{{(1-\alpha )}^2}{1+{\alpha }^2}\right]$

$\therefore$  $f\left(\alpha \right)+f\left({\alpha }^2\right)=f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)$  for all values of $\alpha$  for which the functions are defined, therefore

$\frac{1-\alpha }{1+\alpha }>0\Rightarrow -1<\alpha <\mathrm{1....}\left(1\right)$

$\frac{1-{\alpha }^2}{1+{\alpha }^2}>0\Rightarrow 1-{\alpha }^2>0\Rightarrow -1<\alpha <\mathrm{1....}\left(2\right)$

From (1) and (2), we have $-1<\alpha <1$

$\therefore$  The set of values of$\alpha =(-1,1)$ .