Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for all $x, y\in\mathbb{R}^+$, 
\[ \frac{f(x)}{y^2} - \frac{f(y)}{x^2} \le \left(\frac{1}{x}-\frac{1}{y}\right)^2\]
($\mathbb{R}^+$ denotes the set of positive real numbers.)

[i](Proposed by Ivan Chan Guan Yu)[/i]
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.