c') 0 = ba - gc - dc
a') 0 = gc - da - uay
y') 0 = - ny + uay
There is always an equilibria at {0,0,0}. At that point, all of the animals are dead and obviously will no longer reproduce -- and so will remain dead (at 0).
To solve for the other equilibria, then, assume that neither a, c nor y = 0
y' goes to:
ny = uayc' goes to:
n = ua
a = n/u [which is necessarily >0 ; all parts of it are positive]
ba = gc + dca' goes to:
bn/u = c (g + d)
bn c = --------- [which is necessarily >0 ] u (g + d)
gc - da = uay(which is necessarily > 0 because bg - d(g + d) > 0 by definition in section 2)
(gc - da)/(ua) = y
bg - d(g + d) y = ----------------- > 0 u (g + d)
We can check these on the phase portraits simply by pointing the mouse at the approximate center of the sink on each plot and plugging in the values of {b,g,d,u,n} into the equations for the equilibrium. The center of the sink on the a/c plot should be approximately [1,2/3]; the center on the a/y portrait should be approximately [1,1/3]; and the center on the c/y portrait should be approximately [2/3,1/3].
Note that the other equilibrim (at {0,0,0}) is always a saddle.
So, this system always stabilizes, and it stabilizes at a realistic place -- i.e. the coordinates of the equilibrium are always positive and so all populations are alive -- when bg - d(g + d) > 0, as was set in section 2 to make sure that x didn't automatically die out.
If one allows bg - d(g + d) = 0, then there will be a birfurcation at those parameters; y will eventually die out (y-coordinate of the second equilibrium (the sink) is 0) and x will go to a certain point, as it would with no predation (see Section two)
If one allows bg - d(g + d) < 0, y will be negative at the equilibrium and hence that equilibrium will be irrelevant: y will simply die out. This is because, as seen in section two, x will die out. Then, since the only positive term in y' is related to the prey, y' will always be negative and predators will die out -- in realistic terms, they have no food, so they die.