1. Let $z=x+iy, \quad \bar z=x-iy$ where $x,y\neq 0$.

2. Given equation: $z^2+i\bar z=0$

3. First compute $z^2$: $z^2=(x+iy)^2=x^2-y^2+2ixy$

4. Compute $i\bar z$: $i\bar z=i(x-iy)=ix+y=y+ix$

5. Substitute into the equation: $x^2-y^2+2ixy+y+ix=0$

   Equating real and imaginary parts: $x^2-y^2+y=0 \quad ...(1)$ $2xy+x=0 \quad ...(2)$

6. From (2): $x(2y+1)=0$ Since $x\neq 0$, we get $2y+1=0 \Rightarrow y=-\frac12$

7. Put this into (1): $x^2-\left(-\frac12\right)^2-\frac12=0$ $x^2-\frac14-\frac12=0$ $x^2=\frac34$

8. Now, $|z|^2=x^2+y^2=\frac34+\frac14=1$

9. Therefore, $|z^2|=|z|^2=1$ using the property $|z^2|=|z|^2$ for modulus.

Hence the required value is $\boxed{1}$

10. Checking options:

• A: $9$ ❌
• B: $\frac14$ ❌
• C: $4$ ❌
• D: $1$ ✅