Find all functions $f:\mathbb{Q}^+\rightarrow\mathbb{Q}^+$ such that $$f(x)+f(y)+2xyf(xy)=\frac{f(xy)}{f(x+y)}$$ for all $x,y\in\mathbb{Q}^+$.
Before this problem, I have solved few similar problems and I noticed that easiest way to solve this type of problems is to substitute specific values of $x$ and $y$ to obtain some values of $f$ or to notice some rule. I firstly tried $x=y=1$. From this, I noticed that $f(2)=\frac14$. Then I tried similar values like $(x,y)=\left\{(2,1),(1,2),(2,2),(x,x),\left(x,\frac1x\right),(x,1),\dots\right\}$ but this didn't give me anything.
Then I come to another idea. In this problem we have one very important constraint: domain and range of $f$ must be positive rational numbers. So, I wrote $f(x)$ as $\frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomial functions. Substituting it to main equation and simplifying I got
$$p(x)p(x+y)q(y)q(xy)+p(y)p(x+y)q(x)q(xy)+2xyp(xy)p(x+y)q(x)q(y)=p(xy)q(x)q(y)q(x+y)$$
Now, for $x=y$ we have that degree of LHS must be equal to degree of RHS becuase this is true for any $x$. Let degree of $p(x)$ be $m$ and degree of $q(x)$ be $n$. Now, degree of LHS will be
$$\max\{m+m+n+2n,m+m+n+2n,2+2m+m+n+n\}=\max\{2m+3n,3n+2m+2\}$$
and degree of RHS will be
$$2m+n+n+n=2m+3n$$
Solving equation
$$\max\{2m+3n,3n+2m+2\}=2m+3n$$
I get $n\ge m+2$. This gives me infinitely many solutions, so I don't know how to prove or disprove that it is true for any $f(x)=\frac{p(x)}{q(x)}$ where degree of $p(x)$ is $m$ and degree of $q(x)$ is $n$ such that $n\ge m+2$. I only proved that one correct solution is for $m=0,n=2,p(x)=1,q(x)=x^2$.
I have two questions:
$1.$ If domain and range of some function are rational numbers, does it always mean we can write it as quotient of two polynomials?
$2.$ Can $f(2)=\frac14$ help me to find other solutions or prove this is the only solution? If not, what is the other way?
Here's an attempt to plough through the problem straight ahead. First of all, let's try and find $f(3)$, substituting $x=2, y=1$:
$$f(2)+f(1)+4f(2) = \frac{f(2)}{f(3)}$$ $$f(1)+\frac54=\frac1{4f(3)}$$ $$f(3)=\frac1{4f(1)+5}$$ Awkward, but workable. Now, let's take a look at $f(4)$ the same way, substituting $x=3, y=1$: $$f(3)+f(1)+6f(3)=\frac{f(3)}{f(4)}$$ $$f(4)=\frac{f(3)}{f(1)+7f(3)}=\frac{\frac1{4f(1)+5}}{f(1)+\frac7{4f(1)+5}} =\frac1{4f(1)^2+5f(1)+7}$$
But we can also try plugging in $x=2,y=2$: $$f(2)+f(2)+8f(4)=\frac{f(4)}{f(4)}$$ $$\frac14+\frac14+8f(4)=1$$ $$f(4)=\frac1{16}$$ So now we have a quadratic equation in $f(1)$: $4f(1)^2+5f(1)+7=16$, or $4f(1)^2+5f(1)-9=0$. By inspection, $f(1)=1$ is a solution to this equation, and then by your method of choice (there are at least three different ones that are worth knowing: the quadratic formula, polynomial division of $4x^2+5x-9$ by $x-1$, and Vieta's formula for the sum of roots) the other root can be shown as $-\frac94\not\in\mathbb{Q}^+$. This shows that $f(1)=1$; now can you work the rest of the problem from here?