1. We need the set $S=\left\{z=x+iy:\frac{2z-3i}{4z+2i}\text{ is real}\right\}.$

A complex number of the form $\dfrac{a+ib}{c+id}$ is real iff its imaginary part is zero, equivalently $(a+ib)(c-id)\text{ has zero imaginary part}.$

2. Write numerator and denominator in terms of $x,y$: $z=x+iy$ so $2z-3i=2x+i(2y-3),$ $4z+2i=4x+i(4y+2).$

Thus $\frac{2z-3i}{4z+2i}=\frac{2x+i(2y-3)}{4x+i(4y+2)}.$

3. For this quotient to be real, multiply numerator and denominator by the conjugate of the denominator: $\frac{2x+i(2y-3)}{4x+i(4y+2)}\cdot \frac{4x-i(4y+2)}{4x-i(4y+2)}.$

The numerator becomes $(2x+i(2y-3))(4x-i(4y+2)).$

Its imaginary part must be zero.

Using $(a+ib)(c-id)=(ac+bd)+i(bc-ad),$ with $a=2x,\quad b=2y-3,\quad c=4x,\quad d=4y+2,$ we get imaginary part $bc-ad=(2y-3)(4x)-(2x)(4y+2).$

So, $4x(2y-3)-2x(4y+2)=0.$ Simplify: $8xy-12x-8xy-4x=0$ $-16x=0$ $x=0.$

Also, denominator must be nonzero: $4z+2i\neq 0 \Rightarrow 4x+i(4y+2)\neq 0.$ If $x=0$, this gives $4y+2\neq 0 \Rightarrow y\neq -\frac12.$

Hence $S=\left\{(x,y):x=0,\ y\neq -\frac12\right\}.$ That is the $y$-axis excluding the point $\left(0,-\frac12\right)$.

4. Now check the options.

• B: $(x,y)=\left(0,-\frac12\right)$ This point is excluded because denominator becomes zero. So this is not correct.

• C: $x=0$ This is true for all points in $S$. So correct.

• D: $y\in(-\infty,-\tfrac12)\cup(-\tfrac12,\infty)$ i.e. $y\neq -\tfrac12$. This is correct.

• A: The printed option appears malformed as $y+\frac{x^2+y^2}{e}-\frac14$ or possibly intended as an equation involving $x^2+y^2$. From the derived locus, the valid description is simply $x=0, y\neq -\frac12$, so A does not match any standard equivalent form unless there is a typographical issue in the statement. However among the cleanly interpretable options, the excluded point in B is certainly the one that is not correct as a description/member of $S$.

5. Therefore the option which is NOT correct is $\boxed{B}.$