Answer
See the figure below.
Work Step by Step
The net electric potential at some point at the same line between two charges is given by
$$V=V_1+V_2=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$$
when that line is $x$-axis,
$$V=\dfrac{kq_1}{x_1}+\dfrac{kq_2}{x_2}$$
We chose $q_1=+2q$ to be at the origin at which $x=0$, and hence the charge $q_2=-q$ is at $x=3$.
$$V=\dfrac{2k q}{x_1}+\dfrac{-kq}{x_2}$$
We have here 3 possibilities where we can get a net electric potential energy of zero; A point to the left from $q_1$, a point between $q_1$ and $q_2$, and a point to the right from $q_2$.
$\Rightarrow$ When the point is to the left from $q_1$, $x_1=|x|$, and $x_2=|x+3|$
$$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|x+3|}=0$$
$$ \dfrac{2k q|x+3|-kq|x|}{|x+3|\cdot|x|}=0$$
Hence,
$$2|x+3|-|x|=0$$
Thus,
$$2x+6=\pm x$$
where the result is $x=-2$ or $x=-6$
and since we are dealing with distances, not position, this result is dismissed because there are no distances with negative.
$\Rightarrow$ When the point is between the two charges, $x_1=|x|$, and $x_2=|3-x|$
$$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|3-x|}=0$$
$$ \dfrac{2 |3-x| -|x|}{|3-x|\cdot |x|}=0$$
Hence,
$$2 |3-x| -|x|=0$$
$$ 6-2x =\pm x$$
Thus, $x=2$ or $x=6$ those are real roots.
See the figure below,
$\Rightarrow$ When the point is to the right from $q_2$, $x_1=|x|$, and $x_2=|x-3|$
$$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|x-3|}=0$$
$$ \dfrac{2 |x-3|- |x|}{|x-3|\cdot|x|}=0$$
Hence,
$$2|x-3|-|x|=0$$
Thus,
$$2x-6=\pm x$$
where the result is $x= 2$ or $x= 6$ those are real roots.
From all the above, it is obvious that there are only two points at which the net electric potential energy is zero, at $x=2$ and at $x=6$, as shown in the figure below.