Let $P(x,y)$ denote the assertion. $P(1,1)$ gives us $2f(1)=\left(f(2)+\frac 12\right)f(1)$. i) $f(1)=0$. $P\left(x, \frac 1x\right)$ gives us $f(x)+f\left(\frac 1x\right)=0$......(1) $P(x,1)$ gives us $f(x)=\left(f(x+1)+\frac 1{x+1}\right)(f(x)-x+1)$ Now, let $a=f(2)$ and $b=f(4)$. $P(2,2)$ gives us $2a=\left(b+\frac 14\right)(b-3)$ and $P\left(\frac 12, \frac 12\right)$ gives us $-2a=-b+\frac 34\Rightarrow b-\frac 34=2a$. Hence, $b-\frac 34=\left(b+\frac 14\right)(b-3)\Rightarrow b=0, \frac{15}4$. If $f(4)=b=0$, then $f(2)=a=\frac {-3}8$. $P(2,1)$ gives us $f(3)=\frac{-2}{33}$ and then $P(3,1)$ gives us a contradiction. Hence, $f(4)=b=\frac{15}4$ and $f(2)=a=\frac 32$. Hence, $f\left(\frac 12\right)=\frac{-3}2$. Now, we will prove that $f\left(\frac ab\right)=\frac ab-\frac ba$ by induction over $a+b$ where $a,b\in\mathbb{Z^+}$. It is true for $a+b=2, 3$. Assume that it is true for $a+b=2,3,\cdots ,k$. Let $u+v=k+1$. Let's prove that $f\left(\frac uv\right)=\frac uv-\frac vu$. If $\frac uv\in \left\{\frac 12, 1, 2\right\}$, then we are done. Otherwise, first assume that $u>v$. Then, $P\left(\frac{u-v}v, 1\right)$ gives us the desired result. If $v>u$, then $P\left(\frac{v-u}u, 1\right)$ and (1) gives us the desired result. Hence, $f(x)=x-\frac 1x$ for all $x\in\mathbb{Q^+}$. Clearly, this function works. ii) $f(1)\neq 0$. Then, by $P(1,1)$, we find that $f(2)=\frac 32$. Let $f(1)=c\neq 0$. $P(2,2)$ gives us $f(4)=-1, \frac{15}4$. $P(2,1)$ gives us $f(3)=\frac 83+2c$. $P(3,1)$ gives us $\frac 83+3c=\left(f(4)+\frac 14\right)\left(\frac 23+2c\right)$. If $f(4)=\frac{15}4$, then we find that $c=0$, which is a contradiction. Hence, $f(4)=-1$. This gives us $c=\frac{-19}{27}$. Hence, $f(3)=\frac{34}{27}$. $P(4,1)$ gives us $f(5)=\frac{61}{270}$. Finally, $P(5,1)$ gives us $f(6)=\frac{-245}{6114}$. Now, $P(2,3)$ gives us a contradiction. So, we are done.