Let $B$ be some bounded region of $\mathbb{R}^3$. Let $$p(x,y,z)=\begin{cases} \exp(xyz) & (x,y,z)\in B \\ 0 & \text{otherwise} \end{cases} $$
Show that $X \not \perp Y | Z$.
Where $p(x, y, z)$ is the joint density function of three random variables $X, Y, Z$, and $X \not \perp Y | Z$ means $X, Y$ are not independent given $Z$ (we use a perpendicular symbol with two vertical lines but I'm not good with LaTeX so let's just use one line).
We were given a clue: we should use the fact that $X \perp Y | Z \iff p(x, y, z) = f(x,z)g(y,z)$ for some two non-negative functions $f, g$ for all $x,y,z$ such that $p(z)>0$. Note again that $p(z)$ denotes the probability density function of $Z$.
I can see why splitting the function $\exp(xyz)$ into a multiplication of two functions doesn't work, but am stumped on how to prove this when $B$ is unknown. Any attempt at computing the marginal density functions doesn't work.