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 $$1 - f^{-1}(a) = \frac{1}{a} \quad \Rightarrow \quad f^{-1}(a) = 1-\frac{1}{a} = \frac{a-1}{a}.$$So $ f^{-1}(a) \times a \times f(a) $ is $$\frac{a-1}{a} \times a \times \frac{1}{1-a} = \boxed{-1}.$$