jmc

Setting $y=x$ in the given functional equation, we get $$f(x+1)=f(x)+1+2x.(\ast )$$Then $$\begin{array}{rl}f(x+2)& =f(x+1)+1+2(x+1)\\ & =f(x)+1+2x+1+2(x+1)\\ & =f(x)+4x+4.\end{array}$$Setting $y=2x,$ we get $$f(x+2)=f(x)+\frac{f(2x)}{f(x)}+4x,$$so $$f(x)+4x+4=f(x)+\frac{f(2x)}{f(x)}+4x.$$Hence, $\frac{f(2x)}{f(x)}=4,$ so $f(2x)=4f(x)$ for all $x\in {\mathbb{Q}}^{+}.$

In particular, $f(2)=4f(1).$ But from $(\ast ),$ $f(2)=f(1)+3.$ Solving, we find $f(1)=1$ and $f(2)=4.$ Then $$f(3)=f(2)+1+2\cdot 2=9.$$Setting $x=3$ and $y=1,$ we get $$f\left(3+\frac{1}{3}\right)=f(3)+\frac{f(1)}{f(3)}+2\cdot 1=9+\frac{1}{9}+2=\frac{100}{9}.$$Then by repeated application of $(\ast ),$ $$\begin{array}{rl}f\left(2+\frac{1}{3}\right)& =f\left(3+\frac{1}{3}\right)-1-2\left(2+\frac{1}{3}\right)=\frac{49}{9},\\ f\left(1+\frac{1}{3}\right)& =f\left(2+\frac{1}{3}\right)-1-2\left(1+\frac{1}{3}\right)=\frac{16}{9},\\ f\left(\frac{1}{3}\right)& =f\left(1+\frac{1}{3}\right)-1-2\cdot \frac{1}{3}=\boxed{\frac{1}{9}}.\end{array}$$More generally, we can prove that $f(x)=x^2$ for all $x\in {\mathbb{Q}}^{+}.$