 Solution

For any $x,z\in \{\pm 1\}$ there is exactly one $y=xz$, so $\mathbb{P}(X=x,Z=z)=1/4=\mathbb{P}(X=x)\mathbb{P}(Z=z)$. Similarly $(Y,Z)$ is independent, and $(X,Y)$ is independent by assumption. They are pairwise independent but not jointly independent, because $XYZ=1$ surely.