Answer: $f(x) = x - \frac{1}{x}$, $\forall x \in \mathbb{Q}^+$.
We will prove several lemmas.

$$f(x) + f(y) = \left( f(x + y) + \frac{1}{x+y} \right) (1 - xy + f(xy)) \quad (*)$$

Lemma 1. $f(1) = 0$.
Proof: By putting $x = y = 1$ to (*) we get $2f(1) = \left(f(2) + \frac{1}{2}\right)f(1)$.
Assume that $f(1) \neq 0$. Then readily $f(2) = \frac{3}{2}$. By putting $x = y = 2$ to (*) we get $3 = 2f(2) = \left(f(4) + \frac{1}{4}\right)(f(4) - 3)$. Therefore, either $f(4) = \frac{15}{4}$ or $f(4) = -1$. By putting $y = 1$ to (*) we get $f(x) + f(1) = \left(f(x+1) + \frac{1}{x+1}\right)(1-x+f(x))$. Taking $x = 2,3,4,5$ in the last equation we get
$$\frac{3}{2} + f(1) = \left(f(3) + \frac{1}{3}\right) \cdot \frac{1}{2} \quad (1)$$
$$f(3) + f(1) = \left(f(4) + \frac{1}{4}\right)(f(3) - 2) \quad (2)$$
$$f(4) + f(1) = \left(f(5) + \frac{1}{5}\right)(f(4) - 3) \quad (3)$$
$$f(5) + f(1) = \left(f(6) + \frac{1}{6}\right)(f(5) - 4) \quad (4)$$
If $f(4) = -1$, (1) and (2) yield $f(1) = -\frac{19}{27}$ and $f(3) = \frac{34}{27}$, and (3) and (4) yield $f(5) = \frac{61}{270}$, $f(6) = -\frac{245}{6114}$. Finally by putting $x = 2, y = 3$ to (*) we get
$$f(2) + f(3) = \left(f(5) + \frac{1}{5}\right)(f(6) - 5).$$
This relationship is not held for values of $f(2), f(3), f(5), f(6)$ found above. Hence the only possibility is $f(4) = \frac{15}{4}$. In this case

we have
$$f(3) + f(1) = 4(f(3) - 2), \quad \frac{3}{2} + f(1) = \left(f(3) + \frac{1}{3}\right) \cdot \frac{1}{2}.$$
Solving last two equations we get $f(1) = 0$. Done.

Lemma 2. $f(2) = \frac{3}{2}$.
Proof: Since $f(1) = 0$, for all $x \in \mathbb{Q}^+$ we have
$$f(x) = \left( f(x+1) + \frac{1}{x+1} \right) (1 - x + f(x)) \quad (5)$$
Taking $x = 2, 3$ in (5), we get
$$f(2) = \left(f(3) + \frac{1}{3}\right)(f(2) - 1), \quad f(3) = \left(f(4) + \frac{1}{4}\right)(f(3) - 2).$$
By putting $x = 2, y = 2$ to (*) we get $2f(2) = \left(f(4) + \frac{1}{4}\right)(f(4) - 3)$.
Last three equations yield a cubic equation in terms of $t = f(4)$ as given below:
$$16t^3 - 32t^2 - 101t - 15 = 0.$$
$t = f(4)$ should be rational as the range of $f$ is rational numbers.
The only rational root of this equation is $t = \frac{15}{4}$. Then it readily
follows that $f(3) = \frac{8}{3}$ and $f(2) = \frac{3}{2}$.

Lemma 3. $f(n) = n - \frac{1}{n}$ for all positive integers $n$.
Proof: $f(1) = 0$ for $n = 1$. Proof for $n \ge 2$ readily follows from (5) by induction over $n$. Done.

Finally we prove that for all $x = \frac{m}{n}$
$$f(m/n) = \frac{m}{n} - \frac{n}{m} \quad (6)$$

Proof will be carried out by induction over $m \ge 1$. In the base case $m=1$ by putting $y = \frac{1}{x}$ to (*) we get that for all $x \in \mathbb{Q}^+$
$$f(x) + f\left(\frac{1}{x}\right) = 0$$
and by Lemma 3 for all positive integers $n$ we obtain the required formula
$$f\left(\frac{1}{n}\right) = \frac{1}{n} - n.$$
Now suppose that (6) is correct for $m$. By putting $x = \frac{m}{n}$, $y = \frac{1}{n}$ to (*) we get
$$f(m/n) + f(1/n) = \left(f\left(\frac{m+1}{n}\right) + \frac{n}{m+1}\right) \left(1 - \frac{m}{n^2} + f(m/n^2)\right). \quad (7)$$
Putting $f(m/n) = \frac{m}{n} - \frac{n}{m}$ and $f(m/n^2) = \frac{m}{n^2} - \frac{n^2}{m}$ to (7) and
simplifying we get the required formula for $m+1$:
$$f\left(\frac{m+1}{n}\right) = \frac{m+1}{n} - \frac{n}{m+1}.$$
We are done.