(x/x-1)-(2x/x^2-1)?

[x / (x - 1)] - [2x / (x² - 1)]
= [x / (x - 1)] - [2x / (x - 1)(x + 1)]
= [x(x + 1) / (x - 1)(x + 1)] - [2x / (x - 1)(x + 1)]
= [(x² + x) / (x - 1)(x + 1)] - [2x / (x - 1)(x + 1)]
= [(x² + x) - 2x] / (x - 1)(x + 1)
= (x² + x - 2x) / (x - 1)(x + 1)
= (x² - x) / (x - 1)(x + 1)
= x(x - 1) / (x - 1)(x + 1)
= x / (x + 1)

Should be shown as -

x / [ x - 1 ] - (2x) / [ (x - 1) (x + 1) ]

x (x + 1) - 2x
-------------------
(x - 1) (x + 1)

x² - x
------------------
(x - 1) (x + 1)

x (x - 1)
------------------
(x - 1) (x + 1)

x
---------
x + 1

(x/x - 1) - (2x/x^2 - 1)
= [x(x + 1) - 2x] / (x^2 - 1)
= x / (x + 1)

　
I assume you want to simplify the expression:

.. x ...... 2x
—— − ——
x−1 .... x²−1

If so, the first step is to write it down correctly:
x/(x−1) − 2x/(x²−1)

The second step is to use a common denominator: (x−1)(x+1)
= x(x+1)/((x−1)(x+1)) − 2x/((x−1)(x+1))

And finally, combine and simplify:
= (x(x+1)−2x) / ((x−1)(x+1))
= (x²−x) / ((x−1)(x+1))
= x(x−1) / ((x−1)(x+1))
= x/(x+1)

The denominator x^2 - 1 factors to (x - 1)(x + 1)
Hence the LH term is multiplied through by 'x + 1'
Hence
x(x+1) / (x^2 - 1) - (2x/(x^2 - 1))
(x^2 + x - 2x) / (x^2 - 1)
(x^2 - x) / (x^2 - 1)
Factor
x(x - 1) / (x - 1)(x + 1)
Cancel down by 'x-1'
Hence
x / (x + 1)