To prove ~P begin by assuming P with the aim to demonstrating a contradiction. Thus use "negation introduction". (Note: Do not assume a double negation if you can avoid it - especially if your first move is to eliminate it.)
To obtain a contradiction from ƎxⱯy Qxy use the quantifier instantiation rules and show that Qba is a contradiction for witness b and arbitrary a (which may possibly be b too).
To obtain a contradiction from Eba <-> ~Eaa, for witness b and arbitrary a, well, note that when a is b you have a contradiction. So your universal elimination is to the existential's witness.
To summarise: We should assume ƎxⱯy (Exy <-> ~Eyy), then assume a witness [b] for the existential, that is Ɐy (Eby <-> ~Eyy), so we may eliminate that universal to the witness, so Ebb <-> ~Ebb, and demonstrate this is contradiction, allowing us to thus discharge the assumptions and finish with negation introduction.
Demonstrating that Ebb <-> ~Ebb is a contradiction is simply a matter of using negation elimination and biconditional eliminations. Assuming Ebb derives a contradiction from the biconditional, thus introducing a negation. ~Ebb likewise derives the needed contradicion from the biconditional.
|_
| |_ Ǝx Ɐy (Exy <-> ~Eyy)
| | |_ [b] Ɐy (Eby <-> ~Eyy)
| | | Ebb <-> ~Ebb
| | | |_ Ebb
| | | | ~Ebb
| | | | ┴
| | | ~Ebb
| | | Ebb
| | | ┴
| | ┴
| ~Ǝx Ɐy (Exy <-> ~Eyy)