International Mathematical Olympiad

To tackle this problem, we want to find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy:

1. $f(xf(y)) = yf(x)$ for all $x, y \in \mathbb{R}^+$.
2. $\lim_{x \to \infty} f(x) = 0$.

### Step-by-step Solution:

1. Substitute Special Values:
- Let $y = 1$ in the functional equation.
$$f(xf(1)) = f(x)$$
This implies that if $f$ is not constant, $xf(1)$ must be equal to $x$.

2. Behavior at Infinity:
- Given that $\lim_{x \to \infty} f(x) = 0$, interpret this with $f(xf(y)) = yf(x)$.
- As $x \to \infty$, $f(xf(y)) \to 0$. For $y \neq 0$, this implies $yf(x) \to 0$ for all $y$ with $f(y) \neq 0$.

3. Explore Constants:
- Consider the possibility $f(x) = \frac{1}{x}$:
$$f(xf(y)) = f\left(x \frac{1}{y}\right) = \frac{1}{\frac{x}{y}} = \frac{y}{x} = yf(x)$$
- The function $f(x) = \frac{1}{x}$ satisfies the condition as:
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{x} = 0$$

4. Uniqueness:
- Assume there was another function $g(x)$ satisfying the conditions. Then following similar reasoning and substitutions, you'd obtain:
$$g(x) = \frac{1}{x}$$
- This implies $f(x) = \frac{1}{x}$ is indeed the only solution that satisfies all the conditions.

Hence, the only function that meets the given conditions is:
$$\boxed{f(x) = \frac{1}{x}}$$