 Solution

Lagrange's theorem for finite groups states that if $H\le G$ and $G$ is finite, thenso $|H|$ divides $|G|$. Indeed, the left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ to each coset.

Applying the theorem to the cyclic subgroup $⟨g⟩$ gives

In $C_3\times C_9$, an element has order dividing three exactly when its first coordinate is arbitrary and its second belongs to the unique subgroup of order three. There are $3\cdot 3=9$ such elements, and after removing the identity the answer is $\boxed{8}$.