\( f ( x ) = \frac { \sqrt { 9 - x ^ { 2 } } } { \sqrt { x ^ { 2 } - 1 } } . \)

**Step 1. Set conditions for the numerator** The numerator is \( \sqrt{9-x^2} \). For the square root to be defined (real), the argument must be non-negative: \[ 9 - x^2 \ge 0. \] Solve for \(x\): \[ x^2 \le 9 \quad \Longrightarrow \quad -3 \le x \le 3. \] **Step 2. Set conditions for the denominator** The denominator is \( \sqrt{x^2-1} \). For the square root to be defined, we must have: \[ x^2 - 1 \ge 0. \] In addition, since it is in the denominator, the expression must be strictly positive (to avoid division by zero): \[ x^2 - 1 > 0. \] Solve for \(x\): \[ x^2 > 1 \quad \Longrightarrow \quad |x| > 1 \quad \Longrightarrow \quad x < -1 \quad \text{or} \quad x > 1. \] **Step 3. Combine the conditions** The function is defined when both conditions are met. That is: - \( -3 \le x \le 3 \) from the numerator, and - \( x < -1 \) or \( x > 1 \) from the denominator. The intersection gives: \[ x \in [-3,-1) \cup (1,3]. \] Thus, the domain of the function is \[ \boxed{[-3,-1) \cup (1,3]}. \]