Find all $f : \mathbb{R}\to \mathbb{R}$ , bounded in $(0,1)$ and satisfying:

$x^2 f(x) - y^2 f(y) = (x^2-y^2) f(x+y) -xy f(x-y)$

for all $x,y \in \mathbb{R}$

[i]Proposed by Nikolay Nikolov[/i]
**
   Multiply by 2 and replace \( x \) with \( \frac{x}{2} \):
   \[
   f(x+2) = \frac{x^2 + 8x - 4}{(x+2)^2} f(x)
   \]
   For \( x \neq -2 \).

8. **Combining Results:**
   For \( x \neq -1, -2 \):
   \[
   f(x+2) = \frac{f(x+2)}{f(x+1)} f(x+1) = \frac{(x^2 + 4x - 1)(x^2 + 6x + 4)}{(x+1)^2 (x+2)^2} f(x)
   \]
   This implies \( f(x) = 0 \) for all \( x \neq -1, -2, -\frac{5}{7} \).

9. **Conclusion:**
   Since \( f(x) \) is bounded and satisfies the functional equation, we conclude that \( f(x) = 0 \) for all \( x \). Therefore, the final solution is:
   \[
   f(x) = f(1)x
   \]
   Given \( f(1) = 0 \), we have \( f(x) = 0 \).

The final answer is \( \boxed{ f(x) = 0 } \).