We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x² – 1 ≥ 0

⇒ x² – 1² ≥ 0

⇒ (x + 1)(x – 1) ≥ 0

⇒ x ≤ –1 or x ≥ 1

∴ x ∈ (–∞, –1] ∪ [1, ∞)

In addition, f(x) is also undefined when x² – 1 = 0 because denominator will be zero and the result will be indeterminate.

x² – 1 = 0 ⇒ x = ±1

Hence, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

∴ x ∈ (–∞, –1) ∪ (1, ∞)

Thus, domain of f = (–∞, –1) ∪ (1, ∞)