If $g(x) = 1/f(x)$ then $g : \mathbb{Q}^+ \to \mathbb{R}^+$ and the given equation becomes
$$(*) \quad g(x+y)g(x)g(y) = g(xy)(g(x) + g(y)).$$
Let $f(1) = c > 0$ i.e. $g(1) = \frac{1}{c}$. From (*) we get $g(x+1) = c g(x) + 1$ and hence $g(2) = 2$, $g(3) = 2c+1$, $g(4) = 2c^2+c+1$, $g(5) = 2c^3+c^2+c+1$, $g(6) = 2c^4+c^3+c^2+c+1$. On the other hand setting $x=2$, $y=3$ in (*) leads us to
$$g(5)g(2)g(3) = g(6)(g(2) + g(3)),$$
which implies
$$4c^5 - 3c^3 - c^2 - c + 1 = 0 \Leftrightarrow (c-1)(c+1)(2c-1)(2c^2+2c+1) = 0.$$
If $c=1$ then $g(x+1) = g(x)+1$. By induction $g(n) = n$ for any $n \in \mathbb{N}$ and moreover $g(x+n) = g(x)+n$ for any $x \in \mathbb{Q}^+$ and $n \in \mathbb{N}$.
By setting $y=n$ in (*) we obtain
$$(g(x) + n)g(x)n = g(nx)(g(x) + n),$$
i.e. $g(nx) = n g(x)$. Setting $x=p/q$ and $n=q$, where $p, q \in \mathbb{N}$ we obtain $g(x) = x$, for any $x \in \mathbb{Q}^+$, i.e. $f(x) = 1/x$, for any $x \in \mathbb{Q}^+$.

If $c = \frac{1}{2}$ then $g(x+1) = \frac{1}{2}g(x) + 1$. Hence
$$g(n) = 2 \quad \text{and} \quad g(x+n) - 2 = \frac{g(x) - 2}{2^n}, \quad x \in \mathbb{Q}^{+}, n \in \mathbb{N};$$
Setting $y = n$ in $(*)$ we obtain
$$2g(x+n)g(x) = g(nx)(g(x) + 2),$$
and it's sufficient to see that these equations imply $g(x) = 2$, for any $x \in \mathbb{Q}^{+}$, i.e. $f \equiv 1/2$.
Finally, the only functions satisfying the given equality are $f \equiv \frac{1}{2}$ and $f \equiv \frac{1}{x}$.