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

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

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

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

${f_0} = {f_3} = {f_6} = …….. = \frac{1}{{1 – x}}$

${f_1} = {f_4} = {f_7} = …….. = \frac{{x – 1}}{x}$

${f_2} = {f_5} = {f_8} = …….. = x$

${f_{100}}\left( 3 \right) = \frac{{3 – 1}}{3} = \frac{2}{3}{f_1}\left( {\frac{2}{3}} \right) = \frac{{\frac{2}{3} – 1}}{{\frac{2}{3}}} =  – \frac{1}{2}$

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

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