Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$ holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

[i]Proposed by Titu Andreescu and Gabriel Dospinescu[/i]
1. **Define the polynomial \( Q(x,y,z) \):**
   \[
   Q(x,y,z) = xyz\left( P(x-y) + P(y-z) + P(z-x) \right) - \left( xP(x) + yP(y) + zP(z) \right)
   \]
   We know that \( Q(x,y,z) = 0 \) whenever \( x+y+z = 2xyz \) and \( x, y, z \neq 0 \).

2. **Express \( Q(x,y,z) \) in terms of \( R(x,y,z) \) and \( S(x,y,z) \):**
   There exist polynomials \( R(x,y,z) \) and \( S(x,y,z) \) such that:
   \[
   R(x,y,z) := (x+y+z) \left( P(x-y) + P(y-z) + P(z-x) \right) - \left( xP(x) + yP(y) + zP(z) \right) = S(x,y,z) \cdot \left(2xyz - (x+y+z) \right)
   \]
   This implies that in \( R(x,y,z) \), if we replace every instance of \( xyz \) by \( \frac{x+y+z}{2} \), then we must eventually obtain the zero polynomial. As a corollary, this implies any term of \( R(x,y,z) \) not divisible by \( xyz \) must cancel out.

3. **Simplify by setting \( z = 0 \) and \( x + y = 0 \):**
   \[
   P(x) + P(-x) = 0 \implies P \text{ is even.}
   \]
   Therefore, \( P(x) = P(-x) \).

4. **Consider the degree of \( P \):**
   If \( \deg P \leq 2 \), it is easy to see the mentioned solutions are the only ones. Assume on the contrary that \( \deg P \geq 4 \).

5. **Analyze the highest degree term:**
   Let \( x^{2d} \) be the largest degree term of \( P \). For convenience, assume \( x^{2d} \) has coefficient 1 (by scaling \( P \) by a constant).

6. **Look at the expansion of \( R(x,y,z) \):**
   Let \( \lambda \) denote the coefficient of \( x^{2d}y \). Note \( x^{2d}y \) is the largest degree term. Due to previous analysis, it must also cancel. This forces \( \lambda = 0 \).

7. **Focus on the terms contributing to \( \lambda \):**
   Consider:
   \[
   xP(x-y), \quad yP(x-y), \quad yP(z-x)
   \]
   in the expansion of \( R(x,y,z) \). These are precisely the terms which contribute to \( \lambda \). We obtain:
   \[
   \lambda = -2d + 2
   \]
   Since \( \lambda = 0 \), this forces \( d = 1 \), which is a contradiction!

8. **Conclude the form of \( P(x) \):**
   Therefore, the only possible solutions are polynomials of degree at most 2. Given that \( P \) is even, we can write:
   \[
   P(x) = c(x^2 + 3)
   \]
   for any constant \( c \in \mathbb{R} \).

The final answer is \( \boxed{ P(x) \equiv c(x^2 + 3) } \) for any constant \( c \in \mathbb{R} \).