Solution:

Clearly the function $f(x) = 2$ for all $x \in \mathbb{R}^{+}$ satisfies the given functional equation. We will show that this is the only solution.

Lemma 1: For all $x \in \mathbb{R}^{+}$ we have $f(x) \geq 1$.

To prove this, suppose $f(x) < 1$ for some suitable $x$ and set $y = \frac{x}{1 - f(x)}$. Then $x > 0$ and it follows that $y = x + y f(x)$. From (I) we obtain $f(x) f(x + y f(x)) = 2 f(x + y f(x))$ and since $f(x + y f(x)) > 0$ it follows that $f(x) = 2$, contradiction!

Lemma 2: For all $x \in \mathbb{R}^{+}$ we have $f(x) \geq 2$.

To prove this we set $x = y$ in (I) and obtain (II): $f^{2}(x) = 2 f(x + x f(x))$. Let $f\left(x_{1}\right) < 2$ for some suitable $x_{1}$. Then $f\left(x_{1} + x_{1} f\left(x_{1}\right)\right) = \frac{f^{2}\left(x_{1}\right)}{2} < f\left(x_{1}\right)$. With $x_{k+1} = x_{k} + x_{k} f\left(x_{k}\right)$ for $k = 1, 2, \ldots$ we obtain a monotonically decreasing sequence $\left(f\left(x_{1}\right) = a ; \frac{a^{2}}{2} ; \frac{a^{4}}{2^{3}} ; \ldots ; \frac{a^{2 t}}{2^{2 t-1}}, \ldots\right)$ of function values. For $t > \frac{1}{2 - 2 \log_{2} a}$ the values of this sequence are smaller than 1, contradiction!

Lemma 3: $f$ is monotonically increasing.

To prove this, assume there exist $s, \varepsilon > 0$ with $f(s) > f(s + \varepsilon)$. Substituting $x = s$ and $y = \frac{\varepsilon}{f(s)}$ into (I) gives $f(s) f\left(\frac{\varepsilon}{f(s)}\right) = 2 f(s + \varepsilon)$, from which it follows that $f\left(\frac{\varepsilon}{f(s)}\right) < 2$, contradiction!

Lemma 4: If there exists a $z \in \mathbb{R}^{+}$ with $f(z) > 2$, then $f(x) > 2$ holds for all $x \in \mathbb{R}^{+}$.

Again we substitute $x = z$ and $y = \frac{\varepsilon}{f(z)}$ into (I) and have $f(z) f\left(\frac{\varepsilon}{f(z)}\right) = 2 f(z + \varepsilon)$, from which it now follows with Lemma 2 that: $f(z) \leq f(z + \varepsilon)$ for all $\varepsilon > 0$. Hence there exists a $z_{0} \geq 0$ with $f(z) > 2$ for all $z > z_{0}$. Suppose that $z_{0} > 0$. Then with $x = y = z_{0} - \varepsilon > 0$ it follows from (I) that $f(x) f(y) = 4$, and since $x + y f(x) = \left(z_{0} - \varepsilon\right)\left(1 + f\left(z_{0} - \varepsilon\right)\right) = 3\left(z_{0} - \varepsilon\right)$, for sufficiently small $\varepsilon$ we have $3\left(z_{0} - \varepsilon\right) > z_{0}$ and therefore $2 f(x + y f(x)) > 4$, contradiction!

Lemma 5: If $f(x) > 2$ for all $x \in \mathbb{R}^{+}$, then $f$ is injective.

Suppose there existed $s, \varepsilon > 0$ with $f(s) = f(s + \varepsilon)$. We substitute $x = s$ and $y = \frac{\varepsilon}{f(s)}$ into (I) and have $f(s) f\left(\frac{\varepsilon}{f(s)}\right) = 2 f(s + \varepsilon)$, from which it now follows that $f(s) < f(s + \varepsilon)$, contradiction!

Main proof: Because of the symmetry of the left-hand side of (I), we also have $x + y f(x) = y + x f(y)$. For $y = 1$ it follows that $f(x) = (f(1) - 1) x + 1 = m x + 1$. However, by substitution it is easily shown that no linear function can be a solution of (I).