For $x \in R, x\ne 0, x\ne 1,$ let $f_0\left(x\right)=\frac{1}{1-x}$ and $f_{n+1}\left(x\right)=f_0\left(f_n\left(x\right)\right), n=0, 1, 2,\dots ..$ Then the value of $f_{100}\left(3\right)+f_1\left(\frac{2}{3}\right)+f_2\left(\frac{3}{2}\right)$ is equal to :

$f_1\left(x\right)=f_{0+1}\left(x\right)=f_0\left(f_0\left(x\right)\right)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$

$f_2\left(x\right)=f_{1+1}\left(x\right)=f_0\left(f_1\left(x\right)\right)=\frac{1}{1-\frac{x-1}{x}}=x$

$f_3\left(x\right)=f_{2+1}\left(x\right)=f_0\left(f_2\left(x\right)\right)=f_0\left(x\right)=\frac{1}{1-x}$

$f_4\left(x\right)=f_{3+1}\left(x\right)=f_0\left(f_3\left(x\right)\right)=\frac{x-1}{x}$

$\therefore$ $f_0=f_3=f_6=\dots =\frac{1}{1-x}$

$f_1=f_4=f_7=\dots =\frac{x-1}{x}$

$f_2=f_5=f_8=\dots =x$

$f_{100}\left(3\right)=\frac{3-1}{3}=\frac{2}{3} ,$

$f_1\left(\frac{2}{3}\right)=\frac{\frac{2}{3}-1}{\frac{2}{3}}= -\frac{1}{2}$

$f_2\left(\frac{3}{2}\right)=\frac{3}{2}$

$\therefore$ $f_{100}\left(3\right)+f_1\left(\frac{2}{3}\right)+f_2\left(\frac{3}{2}\right)=\frac{5}{3}$