Sum of solutions to functional equation

Given: (1) $f(x) + 2f\left(\frac{1}{x}\right) = 3x$

Replacing $x$ with $\frac{1}{x}$: (2) $f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}$

Multiplying (2) by 2 and subtracting (1): $3f(x) = \frac{6}{x} - 3x \implies f(x) = \frac{2}{x} - x$

Setting $f(x) = 3$: $\frac{2}{x} - x = 3 \implies x^2 + 3x - 2 = 0$

This quadratic equation has real roots since $\Delta = 9 - 4(1)(-2) = 17 > 0$. By Vieta's formulas, the sum of roots is $-3$.