Condition for neutral equilibrium

Do rate of change of force with respect to $x$ $\frac{dF}{dx} = 0$ at the equilibrium position imply that it is the neutral equilibrium position?

No. For instance, the force $F=-kx^3$ has $\frac{dF}{dx}|_{x=0}=0$, but is nevertheless a stable equilibrium.
Neutral equilibrium requires $\frac{dF}{dx}=0$, but $\frac{dF}{dx}=0$ does not imply neutral equilibrium. Higher-order derivatives need to be examined.
$dF/dx=0$ implies that the position is locally neutral. You'd need to look at the higher order derivatives to see if it was globally neutral. For example consider this force distance curve:
This has $F=0$ and $F'=0$ at $x=0$, but it is obviously only locally neutral.