This can be written as $f(x+y)=f(x)+f(y)+2xy$ for positive rational $x, y$. Now, suppose $g(x)=f(x)-x^2$, then $g(x+y)=g(x)+g(y)$ for all positive rational $x, y$. It is direct that $g(nx)=ng(x)$ for natural $n$. Put $y=\frac{x}{n}$ to get $g\left(\frac{x}{n}\right)=\frac{1}{n}g(x)$. So, this implies $g\left(\frac{p}{q}x\right)=\frac{p}{q}g(x)$. If $x=1$,then $g\left(\frac{p}{q}\right)=\frac{p}{q}g(1)$. So, $f(x)=g(x)+x^2=xg(1)+x^2=x(f(1)-1)+x^2=x(x+f(1)-1)$

Putting $y=kx^2,\mathbb{Z}\owns k\geqslant 1$ we get $f\left((k+1)x\right)=f(x)+f(kx)+2kx^2$ Regarding this as a recursion in $f(kx)$ and solving it, we get $f(kx)=kf(x)+k(k-1)x^2\quad(1)$ Thus plugging $x=1$ we get $f(k)=kf(1)+k^2-k\quad(2)$ Plugging $x={m\over k}, \mathbb{Z}\owns m\geqslant 0$ into $(1)$ we get $f(m)=kf\left({m\over k}\right)+m^2-{m^2\over k}$ Using $(2)$, this becomes $mf(1)+m^2-m=kf\left({m\over k}\right)+m^2-{m^2\over k}$ $f\left({m\over k}\right)=\left({m\over k}\right)^2-{m\over k}+f(1){m\over k}$ Thus for all $x\in\mathbb{Q}^+$ we have $\boxed{f(x)=x^2-x+Cx, C\in\mathbb{Q}^+}$. The solution checks against the given equation.