State the given function.

The function to be expressed in partial fractions is:

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

We assume the form:

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

To find $A$ and $B$, multiply both sides by $(x-1)(x+1)$:

$$2 = A(x+1) + B(x-1)$$

To find $A$, set $x=1$:

$$2 = A(1+1) + B(1-1)$$ $$2 = 2A + 0$$ $$2 = 2A$$ $$A = 1$$

To find $B$, set $x=-1$:

$$2 = A(-1+1) + B(-1-1)$$ $$2 = 0 + B(-2)$$ $$2 = -2B$$ $$B = -1$$

Substitute the values of $A$ and $B$ back into the decomposition:

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

The partial fraction decomposition of $f(x)$ is $\boxed{\frac{1}{x-1} - \frac{1}{x+1}}$.