$$f:R \to R$$ defined as

$$f(x) = x - 1$$ and $$g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$$

Now $$fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$$

$$\therefore$$ Domain of $$fog(x) = R - \{ - 1,1\} $$

And range of $$fog(x) = ( - \infty , - 1] \cup (0,\infty )$$

Now, $${d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}$$

$$\therefore$$ $${d \over {dx}}(fog(x)) > 0$$ for $${{2x} \over {(1 - x)(1 + x)}} > 0$$

$$ \Rightarrow {x \over {(x - 1)(x + 1)}} < 0$$

$$\therefore$$ $$x \in ( - \infty , - 1) \cup (0,1)$$

and $${d \over {dx}}(fog(x)) < 0$$ for $$x \in ( - 1,0) \cup (1,\infty )$$

$$\therefore$$ $$fog(x)$$ is neither one-one nor onto.