Official solution
Given,
f1(x) = \[{1 \over x}\]
f2(x) = 1 - x
f3(x) = \[{1 \over {1 - x}}\]
(f2 \[ \cdot \] J \[ \cdot \] f1) (x) = f3(x)
\[ \Rightarrow \] f2 {J(f1(x))} = f3(x)
\[ \Rightarrow \] f2{J (\[{1 \over x}\])} = \[{1 \over {1 - x}}\]
\[ \Rightarrow \] 1 - J(\[{1 \over x}\]) = \[{1 \over {1 - x}}\]
\[ \Rightarrow \] J(\[{{1 \over x}}\]) = 1 - \[{{1 \over {1 - x}}}\]
\[ \Rightarrow \] J (\[{{1 \over { x}}}\]) = \[{{ - x} \over {1 - x}}\] = \[{x \over {x - 1}}\]
Put x inplace of \[{1 \over x}\]
\[ \therefore \] J(x) = \[{{{1 \over x}} \over {{1 \over x} - 1}}\]
= \[{1 \over {1 - x}} = {f_3}\left( x \right)\]