Complicated rational number functional equation: $f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y$ 
Let $\mathbb{Q}^+$ denote the set of positive rational numbers. Let $f : \mathbb{Q}^+ \to \mathbb{Q}^+$ be a function such that
$$f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y$$
for all $x,$ $y \in \mathbb{Q}^+$. Find all possible values of $f \left( \frac{1}{3} \right)$.

If I substitute in $x=y$, then I get $f(x+1)-f(x)=2x+1$. This suggests that $f(x)=x^2$ works, and one possible value of $f \left( \frac{1}{3} \right)$ is $\frac{1}{9}$. Did I miss anything?

From $f(x+1)-f(x)=2x+1$, we can deduce that $f(x)=x^2+f(1)-1$ for all $x\in\mathbb{Z}^+$.
Note that we also have $f(1+\frac x1)=f(1)+\frac{f(x)}{f(1)}+2x$.
Therefore, $f(x+1)=f(x)+2x+1=\dfrac{f(x)}{f(1)}+2x+f(1)$.
Hence, $f(x)\left(1-\dfrac{1}{f(1)}\right)=f(1)-1$ for $x\in\mathbb{Q}^+$.
As $f$ is not constantly zero, $f(1)=1$. So, $f(x)=x^2$ for $x\in\mathbb{Z}^+$.
$f(3+\frac{1}{3})=f(3)+\frac{f(1)}{f(3)}+2(1)=3^2+\frac{1}{3^2}+2=11+\frac19$
$f(2+\frac{1}{3})=f(3+\frac{1}{3})-2(2+\frac13)-1=5+\frac49$
$f(1+\frac{1}{3})=f(2+\frac{1}{3})-2(1+\frac13)-1=1+\frac79$
$f(\frac{1}{3})=f(1+\frac{1}{3})-2(\frac13)-1=\frac19$
