What is the contrapositive of this statement?

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Here is the statement.

if $x = 2$ and $y = 3$, then $xy = 6$.

Since the contrapositive of $P \Rightarrow Q$ is $\neg Q \Rightarrow \neg P$ would the statement just be

if $xy \neq 6$ then $x \neq 2$ or $y \neq 3$

?

Also, is this statement true or false? I am thinking it's false because the $x$ and $y$ can just swap places so $x = 3$ and $y = 2$ and thus it's still $6$.

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Yes, the contrapositive is "If $xy\ne6$ then $x\ne2$ or $y\ne3$". And it is true. To see that, consider the original statement itself (if the statement is true, so is the contrapositive). This is one of the cases where the truth of the contrapositive is less obvious than the truth of the original statement.

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Your statement of the contrapositive is correct. It's a true statement, just like your original one.

There's no conflict in the contrapositive in the $x=3 ; y=2$ case, since in that situation $\neg Q$ is false (since $xy = 6$), and so we don't need to care about the truth value of $\neg P$ (it happens to be true, since both $x \ne 2$ and $y \ne 3$, either of which would have been enough on its own).

$False \Rightarrow True$ is a true statement (so is $False \Rightarrow False$).

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Indeed, they are equivalent, and true. Other equivalents are also available.

$\begin{array}{r:l}(P\wedge R)\to Q&``\textsf{If $x=2$ and $y=3$, then $xy=6$}" \\[1ex]\neg Q\to (\neg P\vee\neg R)&``\textsf{If $xy\neq 6$, then $x\neq 2$ or $y\neq 3$.}" \\[1ex](\neg Q\wedge P)\to\neg R&``\textsf{If $xy\neq 6$ and $x=2$, then $y\neq 3$.}" \end{array}$

I am thinking it's false because the x and y can just swap places so x is 3 and y is 2 and thus it's still 6.

No, $x=3, y=2$ is not a counterexample, because the statement is not "If $x\neq 2$ or $y\neq 3$, then $xy\neq 6$."   It is that if $xy$ is other than $6$, then it cannot be that both $x$ is $2$ and $y$ is $3$ ... at least one of them must have some other value.