Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\alpha x}}{\mathrm{x}+1};\mathrm{x}\ne -1$, Then value of the constant ‘$\alpha$ ’ such that $\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\mathrm{x}$ is

The given function is $f\left(x\right)=\frac{\alpha x}{x+1};x\ne -1$

Consider the composite function 

$\begin{array}{rcl}f\circ f\left(x\right)& =& x\\ f\left(\frac{\alpha x}{1+x}\right)& =& x\\ \frac{\alpha \left(\frac{\alpha x}{1+x}\right)}{1+\frac{\alpha x}{1+x}}& =& x\end{array}$

Simplify 

$\frac{{\alpha }^{2}x}{1+\left(\alpha +1\right)x}=x$

It implies that 

$\left(\alpha +1\right)x^2+\left({\alpha }^2-1\right)x=0$

This is an identiy, so that $\alpha +1=0\Rightarrow \alpha =\boxed{\mathbf{-}\mathbf{1}}$