Determine all polynomials $P(x)$ with real coefficients and which satisfy the following properties:
i) $P(0) = 1$
ii) for any real numbers $x$ and $y,$
\[|y^2-P(x)|\le 2|x|\quad\text{if and only if}\quad |x^2-P(y)|\le 2|y|.\]
1. **Given Conditions and Initial Setup:**
   We are given that \( P(0) = 1 \) and for any real numbers \( x \) and \( y \), the following equivalence holds:
   \[
   |y^2 - P(x)| \le 2|x| \quad \text{if and only if} \quad |x^2 - P(y)| \le 2|y|.
   \]
   We need to determine all polynomials \( P(x) \) that satisfy these conditions.

2. **Analyzing the Equivalence:**
   Let's analyze the given equivalence condition. We start by considering specific values of \( x \) and \( y \).

3. **Case \( x = 0 \):**
   Substitute \( x = 0 \) into the equivalence:
   \[
   |y^2 - P(0)| \le 0 \quad \text{if and only if} \quad |0 - P(y)| \le 2|y|.
   \]
   Since \( P(0) = 1 \), this simplifies to:
   \[
   |y^2 - 1| \le 0 \quad \text{if and only if} \quad |P(y)| \le 2|y|.
   \]
   The inequality \( |y^2 - 1| \le 0 \) implies \( y^2 = 1 \), so \( y = \pm 1 \).

4. **Substituting \( y = 1 \) and \( y = -1 \):**
   - For \( y = 1 \):
     \[
     |P(1)| \le 2.
     \]
   - For \( y = -1 \):
     \[
     |P(-1)| \le 2.
     \]

5. **Case \( y = 0 \):**
   Substitute \( y = 0 \) into the equivalence:
   \[
   |0 - P(x)| \le 2|x| \quad \text{if and only if} \quad |x^2 - P(0)| \le 0.
   \]
   Since \( P(0) = 1 \), this simplifies to:
   \[
   |P(x)| \le 2|x| \quad \text{if and only if} \quad |x^2 - 1| \le 0.
   \]
   The inequality \( |x^2 - 1| \le 0 \) implies \( x^2 = 1 \), so \( x = \pm 1 \).

6. **Substituting \( x = 1 \) and \( x = -1 \):**
   - For \( x = 1 \):
     \[
     |P(1)| \le 2.
     \]
   - For \( x = -1 \):
     \[
     |P(-1)| \le 2.
     \]

7. **General Form of \( P(x) \):**
   From the above analysis, we see that \( P(1) \) and \( P(-1) \) must both be within the range \([-2, 2]\). To satisfy the given equivalence for all \( x \) and \( y \), we hypothesize that \( P(x) \) must be a quadratic polynomial of the form:
   \[
   P(x) = x^2 + c,
   \]
   where \( c \) is a constant.

8. **Determining the Constant \( c \):**
   Since \( P(0) = 1 \), we have:
   \[
   P(0) = 0^2 + c = 1 \implies c = 1.
   \]
   Therefore, the polynomial is:
   \[
   P(x) = x^2 + 1.
   \]

9. **Verification:**
   We need to verify that \( P(x) = x^2 + 1 \) satisfies the given equivalence condition. For any \( x \) and \( y \):
   \[
   |y^2 - (x^2 + 1)| \le 2|x| \quad \text{if and only if} \quad |x^2 - (y^2 + 1)| \le 2|y|.
   \]
   Simplifying both sides:
   \[
   |y^2 - x^2 - 1| \le 2|x| \quad \text{if and only if} \quad |x^2 - y^2 - 1| \le 2|y|.
   \]
   Since \( |y^2 - x^2 - 1| = |x^2 - y^2 - 1| \), the equivalence holds true.

Thus, the polynomial \( P(x) = x^2 + 1 \) is the only polynomial that satisfies the given conditions.

The final answer is \( \boxed{ P(x) = x^2 + 1 } \).