If $f\left(x\right)=\frac{kx}{x+1},x\ne -1$ , then the value of $k$ for which $f\left(f\left(y\right)\right)=y$ is

$f\left(f\left(y\right)\right)=y\Rightarrow f\left[\frac{ky}{y+1}\right]=y\Rightarrow \frac{k\left(\frac{ky}{y+1}\right)}{\frac{ky}{y+1}+1}=y\Rightarrow \frac{k^{2}y}{ky+y+1}=y$ 

$\Rightarrow k^2=ky+y+1$

$\Rightarrow k^2-1=ky+y$

$\Rightarrow \left(k+1\right)\left(k-1\right)=\left(k+1\right)y$

$\Rightarrow \left(k+1\right)\left[k-1-y\right]=0$

$\Rightarrow k+1=0\Rightarrow k=-1$