If  $f{\left(x\right)}^{2}f\left(\frac{1-x}{1+x}\right)=x^3,x\ne -1,1$ and  $f\left(x\right)\ne 0$, then  $\left\{f(-2)\right\}=\_\_\_\_\_\_\_$ Here {x} represents fractional part of x

Replacing x by  $\frac{1-x}{1+x}$  in the equation
 ${\left(f\left(x\right)\right)}^{2}f\left(\frac{1-x}{1+x}\right)=x^3\mathrm{....}\left(i\right)$
We have  ${\left(f\left(\frac{1-x}{1+x}\right)\right)}^{2}f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^2\mathrm{.....}\left(ii\right)$
Solving (i) and (ii), we have
 ${\left(\frac{x^3}{{\left(f\left(x\right)\right)}^2}\right)}^{2}f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^3\Rightarrow \frac{{\left(f\left(x\right)\right)}^3}{x^6}={\left(\frac{1+x}{1-x}\right)}^3\Rightarrow f\left(x\right)=x^2\left(\frac{1+x}{1-x}\right)$
 $f(-2)=4\left(\frac{-1}{3}\right)=-\frac{4}{3}$
 $\left\{f(-2)\right\}=-\frac{4}{3}-[-\frac{4}{3}]=-\frac{4}{3}+2=\frac{2}{3}$