jmc

Substituting $f^{-1}(a)$ into the expression for $f$, we get $$f(f^{-1}(a))=\frac{1}{1-f^{-1}(a)}.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$a=\frac{1}{1-f^{-1}(a)},$$Solving for $f^{-1}(a)$, we find $$1-f^{-1}(a)=\frac{1}{a}\Rightarrow f^{-1}(a)=1-\frac{1}{a}=\frac{a-1}{a}.$$So $f^{-1}(a)\times a\times f(a)$ is $$\frac{a-1}{a}\times a\times \frac{1}{1-a}=\boxed{-1}.$$