In getting toward (any kind of) disjunctive normal form, you want to start by getting rid of any symbol that isn't a negation, conjunction, or disjunction. The usual rules that you are allowed to use for this are that P ⇒ Q is equivalent to ¬P ∨ Q and that P ⇔ Q is equivalent to (P∧Q)∨(¬P∧¬Q).
Using this you can transform your given sentence into
(¬Y∨(X∨Y∧¬Z)) ∧ ¬((Z∧(X∧¬Y))∨(¬Z∧¬(X∧¬Y)))
I'll note that I am grouping sentences together, according to the convention that the biconditional is the weakest binding operator, followed by conditional, disjunction, conjunction, negation. Some classes do not use this convention, so if yours doesn't, then to be sure that I'm doing what your class requires I'd need to know the binding precedence of your operators.
Since there is some ambiguity about whether your classes uses the binding precedence that I'm used to, I won't go into much more detail than this, since if you use different rules then the details of my answer won't help in your class.
But whatever the precedence rules are in your class, you will still do the same basic first step of exchanging the conditional and biconditional operators for sentences that contain only negation, conjunction, disjunction.
After that step, you want to use De Morgan's laws to distribute the negations into these expressions until negations only occur on the atomic sentences.
From there what you do next depends on the sentence. The sentence might already be in DNF if you're lucky, and then the problem is finished. If you're not lucky, the main connective of the sentence is a conjunction. If so, you'll use your distribution rules to distribute all conjunctions into disjunctions, and the resulting sentence will then be in DNF.
This DNF may not be "full", though. However, just as there are different precedence rules, some classes use different definitions of fullness. So to be sure about how to correctly answer the question, I would want to first check what definition your class uses for "full DNF".
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I want to also note that this is just one way to find the DNF. There is another procedure which uses the truth-table of the sentence. In this method: First draw the truth-table of the sentence. Next, circle every row of the truth-table at which the sentence is true. (Again, I won't do the details of this because the truth-table will depend on your precedence rules, and if your class uses different precedence rules than mine, it might just be a waste of time.)
Now consider any row of the table at which the sentence is true. If X was assigned true in that row, write X followed by a conjunction. If X was assigned false in that row, write ¬X followed by a conjunction. Next do the same for Y: If it was assigned true then write Y and if it was assigned false then write ¬Y, followed by a conjunction. Do likewise next for Z although since this is the end, you won't write a conjunction this time.
What you just constructed will be one disjunct in the DNF. If you do this for every row that you circled, you'll have all the disjuncts, so then you just write a disjunction between them, and the result is the fully DNF.
