Find all positive $x,y\in\Bbb{Q}$ such that $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural numbers. What I did is, the product is also a natural number.
$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$
Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$
Hence $2x$ and $2y$ are natural numbers.
I can't proceed further. Is my way correct? Any help please.

If $x,y$ are rational $x = \frac pq$ and $y = \frac ab$ (assume in lowest terms) so
$\frac pq + \frac ba = \frac {pa + qb}{aq}\in \mathbb Z$
And $\frac ab +\frac qp = \frac {pa+qb}{pb}\in \mathbb Z$
So $q|pa+qb$ so $q|pa$ but $q$ and $p$ are relatively prime so $q|a$.  Likewise $a|pa + qb$ so $a|qb$ so $a|q$ so $a= q$.  And similarly $p = b$.
So $x = \frac 1y$.
So we must have $x + \frac 1y = x + x = 2x$ and $y + \frac 1x = 2y$ are natural numbers.
Whoo boy.
Okay. Case one $x = 1 \in \mathbb Z$ then $y = \frac 11 = 1$ and $x = y = 1$ are obviously solutions.
Case 2: $x \in \mathbb Z$ but $x > 1$.  Then $y=\frac 1x \not \in \mathbb Z$. But $2y=\frac 2x\in \mathbb N$ so $x|2$ so $x = 2$ and $y= \frac 12$ is a solutions ($2+ \frac 1{\frac 12} = 4$ and $\frac 12 + \frac 12 = 1$).
Case 3: $x\not \in \mathbb Z$ then $2x \in \mathbb Z$ so $x = \frac m2$ for some odd $m$.  But then $y =\frac 2m$ and $2y=\frac 4m$ is an integer so $m|4$.  But $m$ is odd so $m =1$.
So those are the only three possibilities.
$x = y = 1;  x = 2; y =\frac 12$ and $x = \frac 12$ and $y = 2$.
