For x ∈ R, x ≠ 0, x ≠ 1, let f _ 0 (x) = (1)/(1 - x) and f _ n + 1...

${f}_{1}\left(x\right) = {f}_{0 + 1}\left(x\right) = {f}_{0}\left({f}_{0}\left(x\right)\right) = \dfrac{1}{1 - \dfrac{1}{1 - x}} = \dfrac{x - 1}{x}$

${f}_{2}\left(x\right) = {f}_{1 + 1}\left(x\right) = {f}_{0}\left({f}_{1}\left(x\right)\right) = \dfrac{1}{1 - \dfrac{x - 1}{x}} = x$

${f}_{3}\left(x\right) = {f}_{2 + 1}\left(x\right) = {f}_{0}\left({f}_{2}\left(x\right)\right) = {f}_{0}\left(x\right) = \dfrac{1}{1 - x}$

${f}_{4}\left(x\right) = {f}_{3 + 1}\left(x\right) = {f}_{0}\left({f}_{3}\left(x\right)\right) = \dfrac{x - 1}{x}$

$∴$ ${f}_{0} = {f}_{3} = {f}_{6} = \ldots = \dfrac{1}{1 - x}$

${f}_{1} = {f}_{4} = {f}_{7} = \ldots = \dfrac{x - 1}{x}$

${f}_{2} = {f}_{5} = {f}_{8} = \ldots = x$

${f}_{100}\left(3\right) = \dfrac{3 - 1}{3} = \dfrac{2}{3},$

${f}_{1}\left(\dfrac{2}{3}\right) = \dfrac{\dfrac{2}{3} - 1}{\dfrac{2}{3}} = - \dfrac{1}{2}$

${f}_{2}\left(\dfrac{3}{2}\right) = \dfrac{3}{2}$

$∴$ ${f}_{100}\left(3\right) + {f}_{1}\left(\dfrac{2}{3}\right) + {f}_{2}\left(\dfrac{3}{2}\right) = \dfrac{5}{3}$