Well, the uniform distribution on $\{(0,1,1), (1,0,1), (1,1,0)\} $ does not factor like that, since it would imply $$0=f(1,1,1)=\prod_{i, j} f_{i, j}(1,1)\ne 0,$$ a contradiction.
To get a positive example, let the probability of $(0,1,1)$, $(1,1,0)$, and $(1,0,1)$ each be $\frac{1-\epsilon}{3}$, and the probability of each of the other 5 elements of $\{0,1\}^3$ be $\epsilon/5$. Then $$\frac\epsilon{5}=f(1,1,1)=\prod_{i, j} f_{i, j}(1,1)\ge \left(\frac{1-\epsilon}3\right)^3,$$ which is a contradiction for small enough $\epsilon$.