#### Answer

$(-\infty, -1) \cup (-1,1) \cup (1, \infty)$

#### Work Step by Step

We want to determine all real $x$ values for which $f(x)=\frac{2}{1-x^2}$ is defined. We can never divide by 0, so set the denominator of $f$ equal to 0 to find all $x$ values that would force us to divide by 0 if plugged into $f$:
$$1-x^2=0.$$
Since the left side of the equation is a difference of squares, we can factor to obtain:
$$(1-x)(1+x)=0$$.
Setting each factor equal to 0 and solving for $x$ gives $x=1$ and $x=-1$. Since these $x$ values give us 0 in the denominator, these are the only two values for which $f$ is not defined. Therefore, the domain of $f$ is all real numbers except $x=1$ and $x=-1$. This can be written in interval notation as $(-\infty, -1) \cup (-1,1) \cup (1, \infty).$