Solution:
The equation is defined if
$$4 a x > 0,\ 4 a x \neq 1,\ x-3 a > 0,\ x-3 a \neq 1$$
Setting $t = \log_{4 a x}(x-3 a)$ we get the equation $t + \frac{1}{2 t} = \frac{3}{2}$ with roots $t_1 = 1$ and $t_2 = \frac{1}{2}$.
If $\log_{4 a x}(x-3 a) = 1$, then $x_1 = \frac{3 a}{1-4 a}$, $a \neq \frac{1}{4}$.
If $\log_{4 a x}(x-3 a) = \frac{1}{2}$ we obtain the equation $x^2 - 6 a x + 9 a^2 = 4 a x$ with roots $x_2 = 9 a$ and $x_3 = a$.
Hence we have to find all $a$ for which exactly two of the numbers $x_1 = \frac{3 a}{1-4 a}\ (a \neq \frac{1}{4})$, $x_2 = 9 a$ and $x_3 = a$ are different solutions of the given equation. It is clear that $a \neq 0$. We shall consider two cases.

1. Let $a > 0$. Then $x_3 - 3 a = -2 a < 0$, which implies that $x_1$ and $x_2$ have to be the solutions. Since $x_1 = \frac{3 a}{1-4 a}$ satisfies (*), it follows that $4 a x_1 = \frac{12 a^2}{1-4 a} > 0$, $4 a x_1 = \frac{12 a^2}{1-4 a} \neq 1$, $x_1 - 3 a = \frac{3 a}{1-4 a} - 3 a > 0$ and $x_1 - 3 a = \frac{3 a}{1-4 a} - 3 a \neq 1$. This implies that $a < \frac{1}{4}$ and $a \neq \frac{1}{6}$. It is easy to check that if $a \neq \frac{1}{6}$, then $x_1 \neq x_2$ and $x_2$ satisfies (*). So the desired numbers $a$ in this case are
$$a \in \left(0, \frac{1}{6}\right) \cup \left(\frac{1}{6}, \frac{1}{4}\right)$$

2. Let $a < 0$. Then $x_2 - 3 a = 6 a < 0$, which implies that $x_1$ and $x_3$ have to be the two solutions. Since $x_1 = \frac{3 a}{1-4 a}$ again satisfies $(*)$, we find as in the first case that $a \neq -\frac{1}{2}$. Now it is easy to check that $x_1 \neq x_3$ and $x_3$ satisfies $(*)$. Thus
$$a \in \left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{2}, 0\right)$$

Combining both cases, it follows that the answer of the problem is
$$a \in \left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{6}\right) \cup \left(\frac{1}{6}, \frac{1}{4}\right)$$