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If you're asking about the difference quotient alone, you have
f (x) = x ² - 2x
which means
f (x + h) = (x + h)² - 2 (x + h)
so that
(f (x + h) - f (x)) / h = (((x + h)² - 2 (x + h)) - (x ² - 2x)) / h
… = (x ² + 2xh + h ² - 2x - 2h - x ² + 2x) / h
… = (2xh + h ² - 2h) / h
*and you would stop here*.
On the other hand, if you're talking about the derivative (which I believe you are, since your question says f '), take the limit as h approaches 0. In particular, this means h ≠ 0, so there is some cancellation:
(f (x + h) - f (x)) / h = (2xh + h ² - 2h) / h
… = 2x + h - 2
Then as h approaches 0, you're left with
f ' (x) = 2x - 2