Official solution
As fn+1(x) = f0(fn(x))
\[ \therefore \] f1(x) = f0+1(x) = f0(f0(x)) = \[{1 \over {1 - {1 \over {1 - x}}}}\] = \[{{x - 1} \over x}\]
f2(x) = f1+1(x) = f0(f1(x)) = \[{1 \over {1 - {{x - 1} \over x}}}\] = x
f3(x) = f2+1(x) = f0(f2(x)) = \[{1 \over {1 - x}}\]
f4(x) = f3+1(x) = f0(f3(x)) = \[{1 \over {1 - {1 \over {1 - x}}}}\] = \[{{x - 1} \over x}\]
\[ \therefore \] f0(x) = f3(x) = f6(x) = . . . . . = \[{1 \over {1 - x}}\]
f1(x) = f4(x) = f7(x) = . . . . . .= \[{{x - 1} \over x}\]
f2(x) = f5(x) = f8(x) = . . . . . . = x
So, f100(3) = \[{{3 - 1} \over 3}\] = \[{2 \over 3}\]
f1\[\left( {{2 \over 3}} \right)\] = \[{{{2 \over 3} - 1} \over {{2 \over 3}}}\] = - \[{1 \over 2}\]
f2\[\left( {{3 \over 2}} \right)\] = \[{{3 \over 2}}\]
\[ \therefore \] f100(3) + f1\[\left( {{2 \over 3}} \right)\] + f2\[\left( {{3 \over 2}} \right)\]
= \[{2 \over 3}\] - \[{1 \over 2}\] + \[{3 \over 2}\]
= \[{5 \over 3}\]