Solution:

If any variable is $0,$ say $z = 0,$ then $zx = 0 \gt y$ is impossible. So $x, y, z \in \{1, \ldots, 8\}$ are distinct positive integers.

The conditions are symmetric. For distinct values $a \lt b \lt c,$ we have $ac \gt b$ and $bc \gt a$ automatically, so the only real constraint is $ab \gt c.$ When it holds, all $6$ orderings work.

Counting $3$-subsets $\{a, b, c\}$ of $\{1, \ldots, 8\}$ with $ab \gt c$ gives $31$ sets. Multiplying by $6$ orderings yields $6 \cdot 31 = 186.$

Thus, the correct answer is C.