$\displaystyle f(x) = \frac{2}{x } - x$ so i have done $\displaystyle f(f(x)) -x = 0$ so $\displaystyle \frac{2}{\frac{2}{ x} -x} - (\frac{2}{ x} -x) - x = 0 $ How does that equal $\displaystyle \frac{4{x}^{ 2} -4}{x(2-{x}^{ 2} ) } $ ???
Shouldn't it be $\displaystyle f(f^{-1}(x)) - x =0$?
it may be easier to write
$\displaystyle y=\frac{2}{x} - x$
$\displaystyle yx=2 - x^2$
Which is a quadratic, so
$\displaystyle x=\frac{-y \pm \sqrt{y^2 + 8}}{2}$
and hence the inverse is
$\displaystyle f^{-1}(x)=\frac{-x \pm \sqrt{x^2 + 8}}{2}$
(not sure if you saw my previous post before i edited it to include the answer)
using your formula you should get the same answer as i posted:
$\displaystyle f(f^{-1}(x)) -x =0$
$\displaystyle \frac{2}{f^{-1}(x)} - f^{-1}(x) -x =0$
multiple everything by f^-1(x):
$\displaystyle 2 - \left[f^{-1}(x)\right]^2 -xf^{-1}(x) =0$
this is a quadratic in $\displaystyle f^{-1}(x)$. the coefficients (a,b,c) are (-1,-x,2)
Solve with quadratic formula
$\displaystyle f^{-1}(x) = \frac{x \pm \sqrt{x^2 + 8}}{-2} = \frac{-x \pm \sqrt{x^2 + 8}}{2}$