Replacing $x = y$ in i) and ii), we get $f(2x) = 4 f(x)$, $\forall x \in \mathbb{R}^{+}$. Therefore, by induction, we can prove that
$$f\left(2^{n} x\right) = 4^{n} f(x), \quad \forall n \in \mathbb{Z}^{+}, x \in \mathbb{R}^{+}.$$
Now, by substituting $y = 2x$ in i) and ii), we get $10 f(x) \leq f(3x) \leq 9 f(x)$, $\forall x \in \mathbb{R}^{+}$. It follows that $f(x) \leq 0, \forall x \in \mathbb{R}^{+}$. Therefore, from ii), we have
$$x y f(x+y) \leq (x+y) y f(x) \leq x y f(x),$$
or
$$f(x+y) \leq f(x), \quad \forall x, y \in \mathbb{R}^{+}.$$
Next, we rewrite the condition ii) as
$$\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x} + \frac{f(y)}{y}.$$
By substituting $y = 2x, 3x, \ldots, n x\left(n \in \mathbb{Z}^{+}\right)$, we can prove by induction that
$$f(n x) \leq n^{2} f(x), \quad \forall n \in \mathbb{Z}^{+}, x \in \mathbb{R}^{+}. \tag{*}$$
From these inequalities, we get
$$\begin{aligned} \frac{2^{n} f(x)}{x} &= \frac{f\left(2^{n} x\right)}{2^{n} x} = \frac{f\left(\left(2^{n} - n\right) x + n x\right)}{\left(2^{n} - n\right) x + n x} \\ &\leq \frac{f\left(\left(2^{n} - n\right) x\right)}{\left(2^{n} - n\right) x} + \frac{f(n x)}{n x} \\ &\leq \frac{\left(2^{n} - n\right) f(x)}{x} + \frac{n f(x)}{x} = \frac{2^{n} f(x)}{x} \end{aligned}$$
for any positive integer $n$ and any positive real number $x$. Therefore, the equality cases in (*) must hold, i.e.
$$f(n x) = n^{2} f(x), \quad \forall n \in \mathbb{Z}^{+}, x \in \mathbb{R}^{+}.$$
From this, we can easily prove that $f(x) = k x^{2}, \forall x \in \mathbb{Q}^{+}$ (where $k = f(1)$, and $k \leq 0$). And then, by combining this result with $f(x+y) \leq f(x), \forall x, y > 0$, we conclude that $f(x) = k x^{2}, \forall x \in \mathbb{R}^{+}$.
It is easy to check that this function satisfy the given conditions.