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 $\boxed{f(x) = x \text{ or } f(x) = -x}$.