International Mathematical Olympiad

To solve the given functional equation problem, we must find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying

$$xf(y) + yf(x) \leq 2.$$

### Step 1: Analyze the Condition

Given the condition $xf(y) + yf(x) \leq 2$, this must be true for exactly one $y$ for each $x$.

### Step 2: Find a Candidate Function

Assume $f(x) = \frac{1}{x}$.

Substitute this into the inequality condition:

$$xf(y) + yf(x) = x \cdot \frac{1}{y} + y \cdot \frac{1}{x} = \frac{x}{y} + \frac{y}{x}.$$

We seek $y$ such that:

$$\frac{x}{y} + \frac{y}{x} \leq 2.$$

### Step 3: Simplify the Expression

The inequality $\frac{x}{y} + \frac{y}{x} \leq 2$ can be rearranged and simplified:

Multiplying through by $xy$ gives

$$x^2 + y^2 \leq 2xy.$$

This simplifies to:

$$(x-y)^2 \leq 0.$$

Hence, we deduce that $x = y$.

### Step 4: Verify Uniqueness

Since we have $(x-y)^2 \leq 0$, it implies $x = y$ is the only solution permissible.

This verifies that for each $x$, the solution for $y$ is unique, and thus the function $f(x) = \frac{1}{x}$ satisfies the condition exactly for one $y = x$.

### Conclusion

The function that meets the problem’s condition is

$$f(x) = \frac{1}{x}.$$

Therefore, the solution to the problem is:

$$\boxed{f(x) = \frac{1}{x}}.$$