Determine all functions $f: \mathbb{R} \to  \mathbb{R}$ such that the inequality
\[ f(x^2)-f(y^2) \le (f(x)+y)(x-f(y)) \]
holds for all real numbers $x$ and $y$.
1. **Define the assertion \( P(x, y) \):**
   \[
   P(x, y): f(x^2) - f(y^2) \le (f(x) + y)(x - f(y))
   \]

2. **Evaluate \( P(0, 0) \):**
   \[
   P(0, 0): f(0^2) - f(0^2) \le (f(0) + 0)(0 - f(0))
   \]
   \[
   0 \le 0 \implies \text{This is always true.}
   \]

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

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

5. **Combine the results from steps 3 and 4:**
   \[
   f(x^2) \le x f(x) \quad \text{and} \quad f(x^2) \ge x f(x)
   \]
   \[
   \implies f(x^2) = x f(x)
   \]

6. **Show that \( f \) is odd:**
   \[
   f((-x)^2) = (-x) f(-x)
   \]
   \[
   f(x^2) = -x f(-x)
   \]
   \[
   x f(x) = -x f(-x)
   \]
   \[
   \implies f(-x) = -f(x)
   \]

7. **Use the new assertion \( Q(x, y) \):**
   \[
   Q(x, y): f(x) f(y) \le x y
   \]

8. **Evaluate \( Q(x, -y) \):**
   \[
   Q(x, -y): f(x) f(-y) \le x (-y)
   \]
   \[
   f(x) (-f(y)) \le -xy
   \]
   \[
   -f(x) f(y) \le -xy
   \]
   \[
   f(x) f(y) \ge xy
   \]

9. **Combine the results from steps 7 and 8:**
   \[
   f(x) f(y) \le xy \quad \text{and} \quad f(x) f(y) \ge xy
   \]
   \[
   \implies f(x) f(y) = xy
   \]

10. **Set \( y = 1 \):**
    \[
    f(x) f(1) = x
    \]
    \[
    f(x) = \frac{x}{f(1)}
    \]

11. **Determine \( f(1) \):**
    \[
    f(1) f(1) = 1
    \]
    \[
    f(1)^2 = 1
    \]
    \[
    f(1) = 1 \quad \text{or} \quad f(1) = -1
    \]

12. **Find the solutions:**
    \[
    \text{If } f(1) = 1, \text{ then } f(x) = x
    \]
    \[
    \text{If } f(1) = -1, \text{ then } f(x) = -x
    \]

The final answer is \( \boxed{f(x) = x \text{ or } f(x) = -x} \).