Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$
This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.
Theorem: $\;\mathbb Z_{\large mn}\;$ is cyclic and $$\mathbb Z_{\large mn} \cong \mathbb Z_{\large m} \times \mathbb Z_{\large n}$$
if and only if $\;\;\gcd(m, n) = 1.$
This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$
It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.
Good-to-know Corollary/Generalization:
The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and
$$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.
Provided you correctly counted the elements of order$~15$, your answer is correct. You can indeed count cyclic subgroups by counting their generators (elements or order$~n$) and dividing by the number $\phi(n)$ of generators per cyclic subgroup, since every element of order$~n$ lies in exactly one cyclic subgroup of order$~n$ (the one that it generates).
Here is how I would count the elements of order$~15$. By the Chinese remainder theorem one has $\newcommand\Z[1]{\Bbb Z_{#1}}\Z{30}\cong\Z2\oplus\Z3\oplus\Z5$ and $\Z{20}\cong\Z4\oplus\Z5$, so all in all we are dealing with the group $\Z2\oplus\Z4\oplus\Z3\oplus\Z5^2$. To have order $15$, an element must have a trivial (zero) component in $\Z2$ and $\Z4$, in $\Z3$ it must have as component one of the $2$ generators, and it's component in $\Z5^2$ must be any one of the $24$ nonzero elements. Indeed you get $2\times24=48$ elements of order$~15$.
Best Answer
Regarding your thoughts on the question:
$$\mathbb Z_{48}\not\cong \mathbb Z_6 \times \mathbb Z_8 \quad \text{and}\quad \mathbb Z_6 \times \mathbb Z_8 \quad \text{is NOT cyclic}.$$
$$\mathbb Z_m\times \mathbb Z_n \;\text{ is cyclic and}\;\; \mathbb Z_m\times\mathbb Z_n \cong \mathbb Z_{mn} \; \text{if and only if}\;\; \gcd(m,n) = 1$$
As you can see, $\gcd(6, 8) = 2 \neq 1$, hence, although $\mathbb Z_6\times \mathbb Z_8$ is abelian, it is not cyclic.
Indeed, $\quad\dfrac{48}{\gcd(6, 8)}= \dfrac {48}{2} = 24.\;$ Note also that $\;\mathbb Z_6 \cong \underbrace{\mathbb Z_2\times \mathbb Z_3}_{\gcd(2, 3) = 1},\;$ so $$\mathbb Z_6\times \mathbb Z_8 = \mathbb Z_2\times \underbrace{\mathbb Z_3\times \mathbb Z_8}_{\gcd(3, 8) = 1} \cong \mathbb Z_2 \times \mathbb Z_{24}$$
So the correct answer, as you state earlier is that the order of the largest cyclic subgroup of $\mathbb Z_6\times \mathbb Z_8$, is the least common multiple of the factor groups: $$\text{lcm}\,(6, 8) = 2^3 \cdot 3 = 24$$
This cyclic subgroup is $\langle (1, 1)\rangle$, the cyclic subgroup generated by $(1, 1)$.