Find all functions $f : R \to  R$ such that $f(x^2)-f(y^2) \le   (f(x)+y) (x-f(y))$ for all $x, y \in  R$.
1. **Initial Setup and Simplification:**
   Consider the functional equation:
   \[
   f(x^2) - f(y^2) \le (f(x) + y)(x - f(y))
   \]
   Let \( P(x, y) \) denote the assertion of the above inequality.

2. **Evaluate \( P(0, 0) \):**
   \[
   P(0, 0) \implies f(0^2) - f(0^2) \le (f(0) + 0)(0 - f(0)) \implies 0 \le -f(0)^2
   \]
   Since \( f(0)^2 \ge 0 \), we must have:
   \[
   0 \le f(0)^2 \le 0 \implies f(0) = 0
   \]

3. **Evaluate \( P(x, 0) \):**
   \[
   P(x, 0) \implies f(x^2) - f(0) \le (f(x) + 0)(x - f(0)) \implies f(x^2) \le x f(x)
   \]

4. **Evaluate \( P(0, y) \):**
   \[
   P(0, y) \implies f(0) - f(y^2) \le (f(0) + y)(0 - f(y)) \implies -f(y^2) \le -y f(y) \implies f(y^2) \ge y f(y)
   \]

5. **Combine Results from Steps 3 and 4:**
   From \( f(x^2) \le x f(x) \) and \( f(x^2) \ge x f(x) \), we get:
   \[
   f(x^2) = x f(x)
   \]
   Let this be \( Q(x) \).

6. **Evaluate \( Q(-x) \):**
   \[
   Q(-x) \implies f(x^2) = -x f(-x)
   \]
   Since \( f(x^2) = x f(x) \), we have:
   \[
   x f(x) = -x f(-x) \implies f(x) = -f(-x)
   \]
   This shows that \( f \) is an odd function.

7. **Expand the RHS in \( P(x, y) \):**
   \[
   f(x^2) - f(y^2) \le (f(x) + y)(x - f(y))
   \]
   Substitute \( f(x^2) = x f(x) \) and \( f(y^2) = y f(y) \):
   \[
   x f(x) - y f(y) \le (f(x) + y)(x - f(y))
   \]
   Simplify the RHS:
   \[
   x f(x) - y f(y) \le x f(x) - f(x) f(y) + xy - y f(y)
   \]
   Cancel terms:
   \[
   0 \le -f(x) f(y) + xy \implies f(x) f(y) \le xy
   \]

8. **Evaluate \( T(-x, y) \):**
   \[
   T(-x, y) \implies f(-x) f(y) \le -xy \implies -f(x) f(y) \le -xy \implies f(x) f(y) \ge xy
   \]
   Combining both inequalities:
   \[
   xy \le f(x) f(y) \le xy \implies f(x) f(y) = xy
   \]

9. **Let \( y = 1 \):**
   \[
   f(x) f(1) = x \implies f(x) = \frac{x}{c} \text{ where } c = f(1)
   \]

10. **Evaluate \( P(1, 1) \):**
    \[
    P(1, 1) \implies 1 \ge f(1)^2
    \]

11. **Evaluate \( P(1, -1) \):**
    \[
    P(1, -1) \implies 0 \le f(1)^2 - 1 \implies f(1)^2 \ge 1
    \]

12. **Combine Results from Steps 10 and 11:**
    \[
    f(1)^2 = 1 \implies f(1) = \pm 1
    \]
    Thus, \( f(x) = x \) or \( f(x) = -x \).

In conclusion, the solutions to the functional equation are:
\[
\boxed{f(x) = x \text{ and } f(x) = -x, \forall x \in \mathbb{R}}
\]