How does energy remain constant when the NET work done is zero?

If I were to go from X to Y and then back to X, the NET work done by me would be zero.

You need to be careful here. Forces, not objects, do work. Additionaly, the total work done on you is equal to your change in kinetic energy. Therefore, a more precise thing to say would be

If I go from X to Y and back to X such that I start and end at rest at X then the net work done on me would be zero.$^*$

If you are talking about walking, then typically accelerations are caused by friction between the ground and your feet. So ultimately if you have started and stopped at rest then friction has done no net work on you. This makes sense. As you speed up, friction points in the direction of your motion, which is positive work being done. As you slow down the opposite is true. Combining these gives no net work.$^{**}$

The whole "energy is the ability to do work" can be a good qualitative picture, but it's usually not as ambiguous when you instead focus on the more precise definitions and relations.

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$^*$ We don't really need to return to the starting point here to get at the heart of the discussion, but I have left it in since it's your scenario.

$^{**}$ This is actually not entirely correct, since, if you aren't slipping, the point of contact between your feet and the ground doesn't move as you take a step. Therefore, friction isn't actually doing any work here. However, if one is more careful with breaking up the various body parts, moving pieces, forces, etc. then one arrives at the same conclusions anyway. I'm choosing a less precise description in order to have a clearer answer.

Ah yes, I think this question can only be answered by the second law of thermodynamics. The second law of thermodynamics states implicitly makes a statement that you can never have a 100% efficient cyclic process. If you first take the body from point A to point B , then you'd have to do some more work to suddenly reverse the process back into the original state.

So, in reality, it is not actually possible to loop back the body back from point A to point B, there will definitely be ton of losses such as friction losses, air resistance losses etc. The amount of loss you can only figure out by measuring the initial and final energy.

You may find this stack exchange interesting, even though the answer talks about thermodynamics, it applies equally well to the example you have involving forces.

But, don't we utilize energy in moving from one point to another and then back?

Actually no. You don't even need to go back. All you have to do is start from $A$ at rest and end at $B$ at rest.

In the absence of dissipation (like air resistance finally transformed into "heat" raising the temperature of air), you don't consume energy at all. You just convert it into motion (called kinetic energy) when you accelerate and get it back when you slow down.

Have a look at power indicator on an eletric car. When you slow down, the car is capable of converting some of the kinetic energy back into electricity and store it in the battery. If things were perfect, it could recover 100% of the energy spent for acceleration when you deccelerate.

Of course, things are FAR from perfect in real life because of many physical processes like friction, air resistance, heat lost in chemical reactions...