Stability of circular orbit in attractive inverse cube central force field

For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$

We can find the potential by $$V=-\int Fdr$$

Hence $$V=-\frac {k} {2r^2}$$ so $$V_{eff}=\frac {l^2} {2mr^2}-\frac {k} {2r^2}$$

and at $r=r_0$, $dV_{eff}/dr=0$ hence

$$dV_{eff}/dr=\frac {-l^2} {mr_0^{3}}+\frac {k} {r_0^3}=0$$

so we have,

$$l^2=mk$$

Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$

$$d^2V_{eff}/dr^2=\frac {3l^2} {mr_0^{4}}-\frac {3k} {r_0^4}$$ Using the above relationship we find that,

$$d^2V_{eff}/dr^2=\frac {3k} {r_0^{4}}-\frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.