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  $\left\{f\left(-2\right)\right\}=$

Here $\left\{x\right\}$ represents the fractional part of $x$.

Replacing x  by$\frac{1-x}{1+x}$ in the equation

 ${\left(f\left(x\right)\right)}^{2}f\left(\frac{1-x}{1+x}\right)=x^3 ...\left(i\right)$

We have 
${\left(f\left(\frac{1-x}{1+x}\right)\right)}^{2}f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^3 ...\left(ii\right)$

Solving (i) and (ii), we have 
${\left(\frac{x^3}{{\left(f\left(x\right)\right)}^2}\right)}^{2}f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^3$

$\Rightarrow \frac{{\left(f\left(x\right)\right)}^3}{x^6}={\left(\frac{1+x}{1-x}\right)}^3$

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

Consider

$\begin{array}{rcl}f\left(-2\right)& =& 4\left(\frac{-1}{3}\right)\\ & =& -\frac{4}{3}\end{array}$

We know that $\left\{f\left(x\right)\right\}=f\left(x\right)-\left[f\left(x\right)\right]$

Hence,

$\left\{f\left(-2\right)\right\}=-\frac{4}{3}-\left[-\frac{4}{3}\right]=-\frac{4}{3}+2=\boxed{\frac{\mathbf{2}}{\mathbf{3}}}$