Suppose there are two events $A$ and $B$ and that $P(A|A\cup B)P(B|A\cup B) = P(A\cap B | A \cup B)$. Then I am asked to find if $A$ and $B$ are independent, positively or negatively correlated.
My intuition is that $A$ and $B$ are independent. If they are,$P(A|A\cap B)P(B|A\cup B)$ would yield only $P(A)P(B)$, and the right side would be equal. But I am not sure if this is a rigorous proof.
What you have is conditional independence given the union of events. (Think about what that says†.)
Independence requires $\mathsf P(A\cap B)=\mathsf P(A)~\mathsf P(B)$ That is all; nothing else.
Total Probability says: $$\begin{align}\mathsf P(A\cap B)~=& ~\mathsf P(A\cap B\mid A\cup B)~\mathsf P(A\cup B) + \mathsf P(A\cap B\mid (A\cup B)^\complement)~\mathsf P((A\cup B)^\complement) \\[1ex] = & ~ \mathsf P(A\cap B\mid A\cup B)~\mathsf P(A\cup B) + 0 \end{align}$$
With what we are given, that means:
$$\begin{align} \mathsf P(A\cap B) ~ = & ~ \mathsf P(A \mid A\cup B)~\mathsf P(B\mid A\cup B)~\mathsf P(A\cup B) \\[1ex] = & ~ \mathsf P(A\mid A\cup B)~\mathsf P(B) \\[1ex] = & ~ \frac{\mathsf P(A)}{\mathsf P(A\cup B)}~\mathsf P(B) \end{align}$$
So your given sets will only be independent if: $\mathsf P(A\cup B)=1$, otherwise $\mathsf P(A\cap B)> \mathsf P(A)~\mathsf P(B)$
† Also consider that $\mathsf P(A\cup B\mid A\cup B) = \mathsf P(A\mid A\cup B)+\mathsf P(B\mid a\cup B)-\mathsf P(A\cap B\mid A\cup B)$