Let $f: \mathbb{Q}^{+} \to \mathbb{Q}^{+}$ be a function such that:
$$f(x) + f(y) + 2xy f(xy) = \frac{f(xy)}{f(x+y)}$$
for all $x, y \in \mathbb{Q}^{+}$.

First, we denote the assertion of the given functional equation as $P(x, y)$.

1. From $P(1, 1)$, we have:
$$f(1) + f(1) + 2 \cdot 1 \cdot 1 \cdot f(1) = \frac{f(1)}{f(2)}$$
$$4f(1) = \frac{f(1)}{f(2)}$$
$$f(2) = \frac{1}{4}$$

2. From $P(n, 1)$, we get:
$$f(n) + f(1) + 2n f(n) = \frac{f(n)}{f(n+1)}$$
$$f(n+1) = \frac{f(n)}{(2n+1)f(n) + f(1)}$$

3. Using the above recurrence relation, we find:
$$f(3) = \frac{f(2)}{(2 \cdot 2 + 1)f(2) + f(1)} = \frac{\frac{1}{4}}{5 \cdot \frac{1}{4} + f(1)} = \frac{1}{5 + 4f(1)}$$
$$f(4) = \frac{f(3)}{(2 \cdot 3 + 1)f(3) + f(1)} = \frac{\frac{1}{5 + 4f(1)}}{7 \cdot \frac{1}{5 + 4f(1)} + f(1)} = \frac{1}{7 + 5f(1) + 4f(1)^2}$$

4. From $P(2, 2)$, we have:
$$f(2) + f(2) + 2 \cdot 2 \cdot 2 \cdot f(4) = \frac{f(4)}{f(4)}$$
$$2f(2) + 8f(4) = 1$$
$$2 \cdot \frac{1}{4} + 8f(4) = 1$$
$$8f(4) = \frac{1}{2}$$
$$f(4) = \frac{1}{16}$$

5. Equating the two expressions for $f(4)$, we get:
$$\frac{1}{7 + 5f(1) + 4f(1)^2} = \frac{1}{16}$$
$$7 + 5f(1) + 4f(1)^2 = 16$$
$$4f(1)^2 + 5f(1) - 9 = 0$$
$$f(1) = 1$$

6. Substituting $f(1) = 1$ into the recurrence relation, we get:
$$f(n+1) = \frac{f(n)}{(2n+1)f(n) + 1}$$

By induction, we can show that $f(n) = \frac{1}{n^2}$ for all $n \in \mathbb{N}$.

7. Now, we claim that $f(x) = \frac{1}{x^2}$ for all $x \in \mathbb{Q}^{+}$.

Using the functional equation and induction, we can extend this result to all positive rational numbers.

Thus, the function that satisfies the given functional equation is:
$$f(x) = \frac{1}{x^2}$$

The answer is: \boxed{\frac{1}{x^2}}.