From your final comment let $p=3$ and $q=13$. This resolves the fact that there is a single subgroup of order $13$ which is normal, and $13$ subgroups of order $3$.
Since the possible orders of subgroups are $1, 3, 13, 39$ this leaves only the question of whether there are any normal subgroups of order $3$.
Think about one such - it is cyclic and consists of three elements $1, a, a^2$. Let $b$ be another element of order $3$. $a$ and $b$ together generate a group of order (a) greater than $3$ because it contains the distinct elements $1, a, a^2, b$; and (b) divisible by $3$ because it contains the element $a$ of order $3$. The only possibility is that $a$ and $b$ generate the whole group.
Now if the subgroup generated by $a$ were normal we'd have either:
$b^2ab=a$ (remembering that $b^3=1$) which would make $a$ and $b$ commute and generate a subgroup of order $9$ and this is impossible in a group of order $39$; or
$b^2ab=a^2$, which we can write $aba=b$ whence $abab=b^2\neq 1$. Then $(ab)^3=b^2ab=a^2\neq 1$ and $(ab)^6=1$. And $ab$ would have order $6$ (we eliminated $2$ and $3$ on the way), and this is impossible in a group of order $39$.
This is not the slickest proof, but shows it can be done with elementary calculations and without Sylow. The general proofs are better because they show more insight into the structure - and the non-abelian groups of order $pq$ have a structure it is worth understanding.
You could also work with $c$, an element of order $13$. We have $a^2ca=c^r: r\neq 1$ because the subgroup of order $13$ is normal and the whole group is not cyclic. And that also leads (with care) to a proof that the group generated by $a$ does not have $c$ in its normaliser, and hence can't be normal.
[Note that $r$ is more restricted than shown above]