1. Given the functional inequality:
$$\frac{f(x)}{y^2} - \frac{f(y)}{x^2} \le \left(\frac{1}{x} - \frac{1}{y}\right)^2$$
for all $x, y \in \mathbb{R}^+$.

2. Let's rewrite the inequality in a more convenient form:
$$\frac{f(x)}{y^2} - \frac{f(y)}{x^2} \le \frac{(x - y)^2}{x^2 y^2}$$

3. Multiply both sides by $x^2 y^2$ to clear the denominators:
$$x^2 f(x) - y^2 f(y) \le (x - y)^2$$

4. Consider the case when $x = y$:
$$x^2 f(x) - x^2 f(x) \le 0 \implies 0 \le 0$$
This is always true, so the inequality holds for $x = y$.

5. Now, let's analyze the inequality for $x \neq y$. Rearrange the inequality:
$$x^2 f(x) - y^2 f(y) \le (x - y)^2$$

6. Divide both sides by $x - y$ (assuming $x \neq y$):
$$\frac{x^2 f(x) - y^2 f(y)}{x - y} \le x - y$$

7. As $x$ approaches $y$, the left-hand side approaches the derivative of $x^2 f(x)$ with respect to $x$. Therefore, we consider the function $g(x) = x^2 f(x)$.

8. The inequality suggests that $g(x)$ is differentiable and:
$$g'(x) = \lim_{y \to x} \frac{g(x) - g(y)}{x - y} \le 0$$

9. Since $g(x) = x^2 f(x)$, we differentiate $g(x)$:
$$g'(x) = 2x f(x) + x^2 f'(x)$$

10. For the inequality to hold, $g'(x)$ must be zero:
$$2x f(x) + x^2 f'(x) = 0$$

11. Solving this differential equation:
$$x^2 f'(x) = -2x f(x)$$
$$f'(x) = -\frac{2}{x} f(x)$$

12. This is a separable differential equation. Separate the variables and integrate:
$$\frac{f'(x)}{f(x)} = -\frac{2}{x}$$
$$\int \frac{f'(x)}{f(x)} \, dx = \int -\frac{2}{x} \, dx$$
$$\ln |f(x)| = -2 \ln |x| + C$$
$$\ln |f(x)| = \ln \left( \frac{C}{x^2} \right)$$
$$f(x) = \frac{C}{x^2}$$

13. Since $f(x)$ maps $\mathbb{R}^+$ to $\mathbb{R}^+$, $C$ must be a positive constant.

Conclusion:
The only function $f$ that satisfies the given inequality for all $x, y \in \mathbb{R}^+$ is:
$$f(x) = \frac{C}{x^2}$$
where $C$ is a positive constant.

The final answer is $\boxed{ f(x) = \frac{C}{x^2} }$ where $C$ is a positive constant.