Using field axioms, prove that x-y=x+(-y)?

L.H.S.
= x - y
= x - y + 0               (Identity)
= x - y + [y + (-y)]    (Inverses)
= (x - y + y) + (-y)    (Commutativity)
= (x + 0) + (-y)        (Inverses)
= x + (-y)                 (Identity)
= R.H.S.

Hence, x - y = x + (-y)

Hints: 

Un-anon yourself to get more responses.
If you asked past questions, go finish them by voting BA.

Once all that done, I will be inclined to answer the current question. 

Important note: Mitcatkie's answer has errors; it is wrong. I'm amazed 
that many have thumbed it up. Just goes to show that many here don't have a clue.

Yes, ted has the correct answer. x - y, the "-" is an operator on the two elements x & y. But in (-y),  the "-" is not an operator, it represents the additive inverse. 

By *definition* or, convention,  the symbolism " x - y" means x + (-y) .
The op is asking to prove a definition. Mitcatkie's mistake was in using what the op wanted to "prove":  y - y is not 0 a priori; it is what the op wants to prove. 

The correct proof of the op's question is the following:

x - y = x + (-y) by definition. QED. It has nothing to do with the axioms of a field. 

comment : you are trying to PROVE a definition ????....what does " - " mean in the field ???...the ONLY operations in the complex numbers are " + , addition " and " x , multiplication "......the definition is x + w = 0 means w is the additive inverse of x and we denote this by the symbols " - x  " , then get " sloppy " and write x - x = 0