What would be the function to make a formula false? 
Produce counterexample to show that the following wfs is not logically
  valid:
$$(\exists x)(\forall y)(A (x,y)\land ¬A (y,x)\implies[A (x,x) \iff A(y,y)])$$

I've tried this interpretations  to see that wfs is not valid but none work:
Domain=$\mathbb N, A(x,y)=x<y$
Domain=$\mathbb N, A(x,y)=x=y$
Domain=$\mathbb N, A(x,y)=x>y$
Domain=$\mathbb N, A(x,y)=x$ and y are even/odd
 A: Two HINTS
First, you have the statement: 
$$(\exists x)(\forall y)(A (x,y)\land ¬A (y,x)\rightarrow [A (x,x) \leftrightarrow A(y,y)])$$
(I replaced the $\implies$ with $\rightarrow$ and $\iff$ with $\leftrightarrow$, since the former ones are typically used for meta-logical claims about logical implication and equivalence)
but you want it to be false. OK, so we want:
$$\neg (\exists x)(\forall y)(A (x,y)\land ¬A (y,x)\rightarrow [A (x,x) \leftrightarrow A(y,y)])$$
to be true.  Well, it may help to push the negation inside:
$$\neg (\exists x)(\forall y)(A (x,y)\land ¬A (y,x)\rightarrow [A (x,x) \leftrightarrow A(y,y)]) \Leftrightarrow$$
$$(\forall x)(\exists y)(A (x,y)\land ¬A (y,x)\land \neg [A (x,x) \leftrightarrow A(y,y)]) \tag{*}$$
This is probably a little easier to work with and think about then the original. For example, from this it should be clear that the relation $A$ cannot be symmetric.
Second, if the 'typical' worlds don't work as counterexamples, you could just try to come up with a world that has one or more objects $a$, $b$, $c$, ..., that do or do not stand in some relation $A$ to each other, without even trying to interpret what those objects and what that relation even means.
In fact, to do this somewhat systematically, you could first just consider a world where there is one object $a$, see if we can construct a counterexample, and if not, add more objects as needed. Now, with one object you're not going to be able to satisfy $(*)$, since you would need both $A(a,a)$ and $\neg A(a,a)$.
OK, but what if we add a second object $b$? Well, that's not going to work either: disregarding the biconditional, you need $A(a,b)$, and $\neg A(b,a)$ to satuisfy the statement for $a$, but you also need $A(b,a)$ and $\neg A(a,b)$ to satisfy the statement for $b$, so you get a contradiction there as well.
OK, then what about adding a third? Or maybe a fourth?  Try it and see what happens. Of course, as you get more objects, you may want to start using a little table to keep track of what relations you do or do not have, i.e. something like:
\begin{array}{c|c|c|c|c|}
A&a&b&c&d\\
\hline
a&1&1&0&0\\
\hline
b&0&0&..&..\\
\hline
c&1&..&..&..\\
\hline
d&1&..&..&..\\
\hline
\end{array}
Good luck!
