1. Given circle

   The circle has center at $(0,0)$ and radius $3$, so its equation is $x^2+y^2=9.$

2. Chord subtending angle $\frac{2\pi}{3}$ at the center

   Let $AB$ be such a chord, and let $M$ be its midpoint.

   Since the perpendicular from the center to a chord bisects the chord, $OM \perp AB$.

   In triangle $OAB$:

   • $OA=OB=3$
   • $\angle AOB = \frac{2\pi}{3}$

   The midpoint $M$ of chord $AB$ lies on the angle bisector of $\angle AOB$, so in right triangle $OMA$, $\angle AOM = \frac{1}{2}\angle AOB = \frac{\pi}{3}.$

3. Find $OM$

   In right triangle $OMA$, $OM = OA\cos\left(\frac{\pi}{3}\right)=3\cdot \frac{1}{2}=\frac{3}{2}.$

   So the midpoint of every such chord is at a fixed distance $\frac{3}{2}$ from the origin.

4. Locus of midpoint

   Hence the locus is the circle centered at the origin with radius $\frac{3}{2}$: $x^2+y^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}.$

5. Check options

   • A: $x^2+y^2=\frac{3}{2}$ ❌
   • B: $x^2+y^2=1$ ❌
   • C: $x^2+y^2=\frac{27}{4}$ ❌
   • D: $x^2+y^2=\frac{9}{4}$ ✅

Therefore, the correct option is D.