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Let's first combine the top two fractions in the numerator.

$\frac{x(x+1)-x(x-1)}{x^2-1}=\frac{x^2+x-x^2+x}{x^2-1}=\frac{2x}{x^2-1}$

This is what our fraction looks like now:
$\frac{\frac{2x}{x^2-1}}{\frac{x}{x^2-1}}$

We can multiply top and bottom by $x^2-1$, which cancels to $\frac{2x}{x}=2$

Multiplying both the numerator and the denominator by $(x^2 - 1)$, we have

$\dfrac{\dfrac{x}{x - 1} - \dfrac{x}{x + 1}}{\dfrac{x}{x^2 - 1}} = \dfrac{x(x + 1) - x(x - 1)}{x} = \dfrac{x((x + 1) - (x - 1))}{x} = \dfrac{2x}{x} = 2$

This holds for $x \neq 0$