Change in gravitational potential energy when the work is zero I may be misunderstanding the whole concept , but my doubt is this.
Let us say there is an isolated system comprising of a rock and the Earth. If I was to lift the rock up with a force equal to the force of gravity on the rock so that the rock moves upwards with a constant velocity. Here the net work is zero because the forces balance each other out. The change in kinetic energy is also zero, but there is an increase in the gravitational potential energy of the system. How is it possible to have an increase in the gravitational potential energy when the net work done is zero.
(note: I may be misunderstanding the concept of systems and potential energy, so any help would be really nice)
 A: 
Here the net work is zero because the forces balance each other out.
The change in kinetic energy is also zero, but there is an increase in
the gravitational potential energy of the system.

The net work is not zero unless you bring the rock to rest at the final height $h$. Then the change in kinetic energy will be zero as the rock begins and ends at rest. Otherwise the rock will have kinetic energy at $h$ in addition to its potential energy. The basic principle here is the work energy theorem which states:
The net work done on an object equals its change in kinetic energy.
So if you lift a rock at rest and bring it to rest at height $h$ its change in kinetic energy is zero and therefore the net work done is zero. What happens here is as follows:
You do positive work of $+mgh$ lifting the rock since your force is in the same direction as the displacement of the rock. At the same time, however, gravity does negative work of $-mgh$ because its force is opposite the direction of the displacement of the rock. Since the rock begins and ends at rest, the net work done is zero.
When something does negative work on an object it takes energy away from the object. In this case, gravity takes the energy you supplied the rock due to your positive work and stores it as gravitational potential energy (GPE) of the rock-earth system. The bottom line is your work winds up as GPE of the rock-earth system. It's important to realize that GPE is a system property. The rock by itself does not possess GPE. The rock-earth system does.
Now if instead of bringing the rock to rest at $h$, it still has velocity $v$ at $h$ then from the work energy theorem the net work done equals $+\frac{1}{2}mv^2$. So the rock has KE in addition to the rock-earth system having GPE.
Hope this helps.
A: 
If I was to lift the rock up with a force equal to the force of gravity on the rock so that the rock moves upwards with a constant velocity.

In your example, the stone was initially lying on the ground at rest (zero velocity). Then you say
"I apply force equal to the gravitational force exerted by the Earth on the stone"
By first Newton's law of motion, when the net force on an object is zero the object remains in equilibrium, which means it keeps moving straight at constant velocity (which can be zero). In other words, you cannot lift the stone from the ground by applying force equal to the gravitational force, there must be some extra force to give the stone kinetic energy (velocity). Once you are satisfied with the stone velocity, by removing this extra force the stone will keep moving straight (upwards) at constant velocity.

Here the net work is zero because the forces balance each other out.

Correct. Net work is zero, but you are forgetting about the extra force that must have been applied to the stone to give it some kinetic energy. Once the stone moves at constant velocity, while you apply the force that equals the gravitational force the net work done on the stone is zero.

How is it possible to have an increase in the gravitational potential energy when the net work done is zero.

Net work being zero only means that kinetic energy of the object does not change, which follows directly from the work-energy theorem
$$\Delta K = W$$
where $W$ is total work done by all forces combined. The above equation can be rewritten as
$$\Delta K + \Delta U_g = W_\text{other}$$
where $W_\text{other}$ is total work done by all forces except the gravitational force, and $\Delta U_g$ is change in the gravitational potential energy. Note that work done by the gravitational force is $W_g = -\Delta U_g$ by definition. It is now obvious that if kinetic energy does not change $\Delta K = 0$, the gravitational potential energy changes due to the (upward) force you exert on the stone.
