Answer: C

Small Hint:

None of $x, y, z$ can be $0,$ since then some product equals $0$ and cannot exceed a nonnegative value

Big Hint:

For distinct $a \lt b \lt c,$ the only binding condition is $ab \gt c$; count such $3$-subsets, then multiply by $6$

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.

For each possible smallest value $a,$ the numbers of pairs $a\lt b\lt c\le8$ satisfying $ab\gt c$ are $$\begin{array}{c|rrrrrr} a&1&2&3&4&5&6\\ \hline \text{count}&0&11&10&6&3&1. \end{array}$$ For fixed $a,b,$ this count comes from choosing $c$ between $b+1$ and $\min(8,ab-1).$ Thus there are $31$ unordered triples, and all $6$ orderings of each work. The answer is $6\cdot31=186.$

Thus, the correct answer is C.