A Function that Has Two Roots!

First, substitute $x = a$ into the given equation, then we have

$f(a) + 2f\left( \frac{1}{a} \right) = 3a \quad ...(1)$ .

Second, substitue $x = \dfrac{1}{a}$ into the given equation, then we have

$f\left(\frac{1}{a}\right) + 2f(a) = \frac{3}{a} \Rightarrow f\left( \frac{1}{a} \right) = \frac{3}{a} - 2f(a) \quad ...(2)$ .

Substitute the equation $(2)$ into $(1)$ , we have

$f(a) + 2\left[ \frac{3}{a} - 2f(a) \right] = 3a$

$f(a) + \frac{6}{a} - 4f(a) = 3a \Rightarrow -3f(a) = 3a - \frac{6}{a}$

$f(a) = \frac{2}{a} - a$ .

Since $f(a) = f(-a)$ , then we have $\frac{2}{a} - a = \frac{2}{-a} - (-a)$

$\frac{2}{a} -a = -\frac{2}{a} + a \Rightarrow 2a = \frac{4}{a}$

We have $a^2 = 2 \rightarrow a = \pm \sqrt{2}$ .

Hence, $a = \boxed{(\sqrt{2} , -\sqrt{2})}$ .