If f(x)= then f'(2)= ?

Remember...

$\frac{d}{dx}u^3 \,=\, 3u^2\,\cdot\,\frac{d}{dx}u$

So

 $f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,\frac{d}{dx}[x+\frac1x-1] \\~\\ f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,[\frac{d}{dx}x+\frac{d}{dx}\frac1x-\frac{d}{dx}1] \\~\\ f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,[1-x^{-2}]$

To find  f'(2) , plug in  2  for  x .

$f'(2)\,=\,3[2+\frac12-1]^2\,\cdot\,[1-2^{-2}] \\~\\ f'(2)\,=\,3[\frac32]^2\,\cdot\,[1-\frac1{4}] \\~\\ f'(2)\,=\,3[\frac94]\,\cdot\,[\frac3{4}] \\~\\ f'(2)\,=\,\frac{81}{16}$

( If   f(x)  is meant to be  $[\frac{x+1}{x-1}]^3$  then you get a different answer. )

If  f(x)  =  [x + 1/x - 1]³,  f(x)  is of the form  f(x) = [g(x)]³, so  f'(x)  =  3[g(x)]²·g'(x)

  f(x)  =  [x + 1/x - 1]³     --->     f'(x)  =  3[x + 1/x - 1]³·[1 - 1/x²]

To find f'(2), replace x with 2 in the above derivative, and simplify.