Given,

The complex number $z=x+iy$ be such that $\frac{2z-3i}{2z+i}$ is purely imaginary,

Now putting the value of $z=x+iy$ in $\frac{2z-3i}{2z+i}$ we get,

$\frac{2\left(x+iy\right)-3i}{2\left(x+iy\right)+i}=\frac{\left(2x+i\left(2y-3\right)\right)}{\left(2x+i\left(2y+1\right)\right)}\frac{\left(2x-i\left(2y+1\right)\right)}{\left(2x-i\left(2y+1\right)\right)}$

Now taking real part as,

$\frac{4x^2+\left(2y-3\right)\left(2y+1\right)}{4x^2+{\left(2y+1\right)}^2}=0$

$\Rightarrow 4x^2+4y^2-4y-3=0$

$\Rightarrow x^2+y^2-y-\frac{3}{4}=0$

Now using,

$x+y^2=0$$\Rightarrow x=-y^2$ we get,

$x^2+y^2-y-\frac{3}{4}=0$

$\Rightarrow y^4+y^2-y-\frac{3}{4}=0$

$\Rightarrow y^4+y^2-y=\frac{3}{4}$