Do maxima in Gibbs energy also correspond to equilibrium positions? ΔrG=(∂G/∂ξ ) at constant p and constant T, where ξ is the extent of reaction.

By Second Law implication, ΔrG= 0= (∂G/∂ξ ) at equilibrium. If we consider a graph of G against ξ, this would mean  we have a turning point (where first derivative=0) when ξ is that extent of reaction when equilibrium attained:   


  (∂G/∂ξ )= 0 when ξ= extent of reaction until equilibrium
  attained
  


So, ΔrG=0 when equilibrium attained.

However, first derivative=0 does not mean minimum point only, it can mean maximum point too! Let's say you do work until you reach Gibbs energy maxima. My question: Does a Gibbs energy maxima correspond to equilibrium state or not?

(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease. Just simply move from the maximum point and you will at least be better than your current state:

(b) If no, then it will mean that we have to move from the maximum point, but in which direction? In case the graph is a symmetrical parabola, where same dξ results in same dG, what will you observe in real life? Shift to reactants or products?
G vs. $\xi$ does not have a maximum

If all reactants and products are pure liquids and solids, G vs. $\xi$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.

Analogy to mechanics

In mechanics, this situation would be called metastability. Giving a nudge in either direction will allow the system to reach lower potential energy.