Answer
$\pm 12\;\rm cm$
Work Step by Step
As we see in the figure below, the points at which the net electric potential is zero are $y_P$ and $y_{P'}$ where $y_P=y_{P'}$ according to the symmetry.
Hence,
$$V_{net}=V_1+V_2=0$$
$$\dfrac{ \color{red}{\bf\not} kq_1}{r_1}+\dfrac{ \color{red}{\bf\not} kq_2}{r_2}=0$$
$$\dfrac{ q_1}{r_1}=-\dfrac{ q_2}{r_2} $$
where $r_1=\sqrt{x_1^2+y_p^2}$ and $r_2=\sqrt{x_2^2+y_p^2}$
$$\dfrac{ q_1}{\sqrt{x_1^2+y_p^2}}=-\dfrac{ q_2}{\sqrt{x_2^2+y_p^2}} $$
Plug the known and solve for $y_p$,
$$\dfrac{ -3}{\sqrt{(-9)^2+y_p^2}}=-\dfrac{ 4}{\sqrt{16^2+y_p^2}} $$
Squaring both sides,
$$9(256+y_p^2)=16(81+y_p^2)$$
$$2304+9y_p^2=1296+16y_p^2$$
$$7y_p^2=1008$$
$$y_p^2=144$$
Hence,
$$y_p=\pm 12\;\rm cm$$
So, we have two points at which the electric potential is zero which are
$$y=\color{red}{\bf 12}\;\rm cm$$
and
$$y=\color{red}{\bf -12}\;\rm cm$$
