4. The function is

$$
f(x)=\frac{a x+2}{3 x-\frac{1}{a}}
$$

Determine all possible values of the real parameter $a$ such that for all real values of $x$ for which $f(x)$ is defined, it is also true that $f(f(x))$ is defined and $f(f(x))=x$.
4. Note that the function $f$ is defined for all real arguments except for $\frac{1}{3a}$. Therefore, for $f(f(x))$ to be defined, the equation $f(x) = \frac{1}{3a}$ must have no solutions $x$ in the set $\mathbb{R} \backslash \left\{\frac{1}{3a}\right\}$. The equation $f(x) = \frac{1}{3a}$, i.e., $\frac{ax + 2}{3x - \frac{1}{a}} = \frac{1}{3a}$, simplifies to $3a^2 x + 6a = 3x - \frac{1}{a}$, which is equivalent to $\left(a^2 - 1\right)x = -2a - \frac{1}{3a}$. In the case $a = \pm 1$, it is clear that there are no solutions $x$ (which is what we wanted). For $a \neq \pm 1$, we have $x = \frac{-2a - \frac{1}{3a}}{a^2 - 1}$, and the only case that fits here is when $x$ is $\frac{1}{3a}$, i.e., $\frac{1}{3a} = \frac{-2a - \frac{1}{3a}}{a^2 - 1}$, which simplifies to $a^2 - 1 = -6a^2 - 1$, i.e., $a = 0$, but this is impossible because for $a = 0$ the expression in the problem is not defined.

Therefore, only $a = \pm 1$ remains, and we need to check whether both these values of $a$ satisfy the conditions in the problem. For $a = 1$, we have $f(x) = \frac{x + 2}{3x - 1}$ and $f(f(x)) = \frac{\frac{x + 2}{3x - 1} + 2}{3 \frac{x + 2}{3x - 1} - 1} = \frac{\frac{7x}{3x - 1}}{3x - 1} = x$, which is what we wanted. For $a = -1$, we have $f(x) = \frac{-x + 2}{3x + 1}$ and $f(f(x)) = \frac{-\frac{-x + 2}{3x + 1} + 2}{3 \frac{-x + 2}{3x + 1} + 1} = \frac{\frac{7x}{3x + 1}}{3x + 1} = x$, which is what we wanted.

Therefore, there are two desired values of $a$, and they are $a = 1$ and $a = -1$.