You obtained the correct derivative, but you need parentheses as such: $\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x \frac{1}{2}(x^2-1)^\frac{-1}{2}2x\Bigr)$
Clean this up a bit to get $\tag{1} \frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x^2 (x^2-1)^\frac{-1}{2} \Bigr). $ You were ok up to this point. The rest of your work contains an algebraic error: in the second line of the displayed equations after you say "which I reduce to", the second term is off, it needs an "$x$" downstairs.
But other than that, you did fine. For what it's worth here is the derivation with the correction: Equation ${1}$ can be written as $ \frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr) $ Now multiply through
$\eqalign{ \frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr) &= \frac{1}{x \color{maroon}{\sqrt{x^2-1}}} \cdot \color{maroon}{\sqrt{x^2-1} }+ \frac{1}{\color{darkblue}x\color{darkgreen}{ \sqrt{x^2-1}}} \cdot {\color{darkblue}{x^2}\over\color{darkgreen}{\sqrt{x^2-1}}} \cr &={1\cdot\color{maroon}1\over x} +\frac{\color{darkblue}x}{ \color{darkgreen}{{x^2-1}}} \cr &={{(x^2-1)}\cdot1+x\cdot x\over x(x^2-1)}\cr &={2x^2-1\over x(x^2-1)}.\cr } $