If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations!
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer. The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.
Note that if $(m, n)$ is a solution then $(mn-1) \mid (n^3-1)$ so $(mn-1) \mid m^3(n^3-1)$. Also $(mn-1)\mid m^3n^3-1$ so $(mn-1) \mid m^3-1$, so $(n, m)$ is also a solution.
We may thus WLOG assume $m \geq n$. If $n=1$ then $\frac{n^3-1}{mn-1}=0$ so $(m, 1)$ is a solution for all positive integers $m>1$. I would reject $(1, 1)$ as that leads to $\frac{0}{0}$.
Otherwise $n \geq 2$. Note that $\frac{n^3-1}{mn-1} \equiv 1 \pmod{n}$ so we may write $\frac{n^3-1}{mn-1}=kn+1$, for some $k \in \mathbb{Z}$.
Now $m, n \in \mathbb{Z}^+$ so $0<\frac{n^3-1}{mn-1}=kn+1$. Since $n \geq 2$, this implies $k \geq 0$.
Since $m \geq n$, we have $kn+1=\frac{n^3-1}{mn-1} \leq \frac{n^3-1}{n^2-1}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}<n+1$. Thus $k<1$.
Therefore $k=0$, so $\frac{n^3-1}{mn-1}=1$, so $n^3-1=mn-1$, so (since $m, n>0$) $m=n^2$.
In conclusion, all positive integer solutions are given by $(m, 1)$ for $m>1$, $(1, n)$ for $n>1$, $(m, m^2)$ for $m>1$, and $(n^2, n)$ for $n>1$.
Best Answer
We are looking for $9ab\ |\ 9a^2+14b^2$.
Now, as everything else is divisible by $9$, we immediately get that $9\ |\ b^2$, i.e., $3\ |\,b$.
Similarly, looking at divisibility by $a$, we get $a\,|\,14$. This gives only four possibility on $a$: it can be $\ 1,\ 2,\ 7,\ 14$.
If we write $b=3k$, we get $27ak\ |\ 9a^2+14\cdot 9k^2$, that simplifies to $$3ak\ |\ a^2+14k^2\,.$$
Now, by divisibility by $k$, we have $k\,|\,a^2$, which is only possible if $k=1$ by condition $\gcd(a,b)=1$.
From this there are only $4$ possibilities, check them manually.