Let, $z=x+i y$

So, $\frac{2 z-3 i}{2 z+i}$ is purely imaginary $~~$[Given]

$$ \begin{aligned} & \text { Now, }\left(\frac{2 z-3 i}{2 z+i}\right)+\left(\frac{\overline{2 z-3 i}}{2 z+i}\right)=0 \\\\ & \Rightarrow \frac{2 z-3 i}{2 z+i}+\frac{2 \bar{z}+3 i}{2 \bar{z}-i}=0 \end{aligned} $$

$$ \begin{aligned} & \Rightarrow(2 z-3 i)(2 \bar{z}-i)+(2 \bar{z}+3 i)(2 z+i)=0 \\\\ & \Rightarrow\left\{4|z|^2-6 i \bar{z}-2 i z-3\right\}+\left\{4|z|^2+6 i z+2 i \bar{z}-3\right\}=0 \\\\ & \Rightarrow 8|z|^2-4 i \bar{z}+4 i z-6=0 \\\\ & \Rightarrow 8\left(x^2+y^2\right)-4 i(x-i y)+4 i(x+i y)-6=0 \\\\ & \Rightarrow 8\left(x^2+y^2\right)-4 i x-4 y+4 i x-4 y-6=0 \\\\ & \Rightarrow 8\left(x^2+y^2\right)-8 y-6=0 \\\\ & \Rightarrow 4\left(x^2+y^2\right)-4 y-3=0 \end{aligned} $$

Given that, $x+y^2=0$

$$ \begin{aligned} & \Rightarrow x=-y^2 \\\\ & \Rightarrow 4\left(y^4+y^2\right)-4 y=3 \\\\ & \Rightarrow y^4+y^2-y=\frac{3}{4} \end{aligned} $$

Hence, required answer is $\frac{3}{4}$.