Determine if it is a tautology, contradiction, or contingency I'm trying to figure out is it a tautology, contradiction, or contingency?
$(∀x.∀y.p(x,y))↔(∀y.∀x.p(x,y))$
I think $(∀x.∀y.p(x,y))$ and $(∀y.∀x.p(x,y))$ both true, so the answer is tautology, but my teacher told me I'm wrong.
Who can tell me why?
The truth of the individual statements $\forall x \forall y \ p(x,y)$ and $\forall y \forall x \ p(x,y)$ depends on what $p(x,y)$ is saying and what the domain is over which you are quantifying. For example, if $p(x,y)$ means that $x=y$, then $\forall x \forall y P(x,y)$ is saying that everything is identical to everything ... which would be false for most domains though if your domain contains exactly one object, it would actually be true.

So no, you should not try to evaluate the truth-values of the individual statements, since you cannot know those. 

Also, even if you would know both statements to be true, then that would only tell you that for that one possible interpretation, the biconditional is true, but in order for the biconditional to be a tautology, the biconditional needs to be true in every possible interpretation, not just for one possible interpretation.

So, if you think the biconditional is a tautology, you need to provide a general argument that shows that no matter how we interpret $p(x,y)$, and no matter what domain we use, the left and right side of the biconditional will always have the same truth-value.

If, on the other hand, you think the statement is a contradiction, you need to show how no matter how we interpret $p(x,y)$, and no matter what domain we use, the left and right side of the biconditional will always have the opposite truth-value.

And finally, if you think the statement is neither, then you need to come up with some interpretation in which the two statements have the same truth-value, as well as an interpretation where they have opposite truth-values.
It's the three ones under disjunction, if sets with x, y are given and for all the elements of those sets p(x,y) is true, then ∀x.∀y.p(x,y) is true and implies ∀x.∀y.p(x,y) so your statement (∀x.∀y.p(x,y))↔(∀y.∀x.p(x,y)) is a tautology, if the given sets can't make true p(x,y) for some x, y, then ~(∀x.∀y.p(x,y)) that implies ~(∀x.∀y.p(x,y)) (since it's iff and by modus tollens) so your statament is not a tautology but it's not necessarily a contradiction too, thus what I can say about it it's that your statement it's a contradiction xor your statement it's a tautology xor your statement it's a contigency, in fact nothing special to find.