(I assume that $U$ denotes some universal set, or universe of discourse: simply put, a set which contains everything currently under discussion.)
Well, $A$ union everything must be at least as big as everything. But everything is, well, everything, so no set can be bigger than everything. Therefore $A$ union everything must be everything.
Your first and last $\iff$’s are meaningless, I’m afraid. The statement $P\iff Q$ means that either $P$ and $Q$ are both true, or $P$ and $Q$ are both false. In particular, $P$ and $Q$ have to be statements, things that can be true or false. $(A\cap B)^c$, however, is a set, not a statement: it can no more be true or false than a symphony can be pink.
The statement $\neg\big(P(x)\land Q(y)\big)$ actually says that it is not the case that $x\in A$ and $y\in B$. What are $x$ and $y$? They’ve come out of nowhere. What you want, I think, is to say that an object belongs to $(A\cap B)^c$ if and only if it is not the case that it belongs to $A$ and to $B$:
$$x\in(A\cap B)^c\iff\neg\big(P(x)\land Q(x)\big)\;.$$
Now you can proceed much as you did before, applying one of the logical De Morgan’s laws:
$$\neg\big(P(x)\land Q(x)\big)\iff\neg P(x)\lor\neg Q(x)\;.$$
That step was fine, except that you were working with an expression that wasn’t quite right to begin with. In the last step, however, you have the same problem that you had in the first step: $A^c\cup B^c$ is not a statement, so it makes no sense to connect it to something with $\iff$. What you want here is
$$\neg P(x)\lor\neg Q(x)\iff x\notin A\lor x\notin B\iff x\in A^c\lor x\in B^c\iff x\in A^c\cup B^c\;.$$
And since you have $\iff$ ‘if and only if’ at each step, this argument shows that the elements of $(A\cap B)^c$ are exactly the same as those of $A^c\cup B^c$ and hence that $(A\cap B)^c=A^c\cup B^c$: you’re done at that point.
Best Answer
We have :
$$(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)=((A \cap B) \cup (A'\cap B))\cup (A \cap B \cap C' \cap D)$$
You just use here the associativity and comutatitivity of $\cup$. Now :
$$(A \cap B) \cup (A'\cap B)=(A\cup A')\cap B=U\cap B=B$$
I used distribution of $\cap$ with respect to $\cup$. Then $A\cup A'=U$ because $A\cup A'$ consists by definition of elements of $U$ which are in $A$ or not in $A$ (clearly, all elements of $U$ verify this). Then $U\cap B=B$ because $B\subseteq U$. Now you have :
$$(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)=B\cup (A \cap B \cap C' \cap D)$$
And I will let you finish...