A function f, defined for all non-zero real x, satisfies f(x) + 2f(1/x) = 3x. For how many non-zero real numbers x is f(x) = f(−x)?

Replace x by 1/x: f(1/x) + 2f(x) = 3/x.
Solve with the original equation: multiply the new one by 2 and subtract the original: 4f(x) − f(x) = 6/x − 3x ⇒ f(x) = 2/x − x.

Then f(−x) = −2/x + x = −f(x), so f is odd.
f(x) = f(−x) ⇒ f(x) = −f(x) ⇒ f(x) = 0 ⇒ 2/x = x ⇒ x² = 2 ⇒ x = ±√2.

So there are 2 such values.

Check: f(2) = 1 − 2 = −1, f(1/2) = 4 − 1/2 = 3.5; f(2) + 2f(1/2) = −1 + 7 = 6 = 3 × 2 ✓.