6. Find all real-coefficient polynomials $P$ such that for all non-zero real numbers $x, y, z$ satisfying $2xyz = x + y + z$, we have
$$
\begin{array}{l}
\frac{P(x)}{yz} + \frac{P(y)}{zx} + \frac{P(z)}{xy} \\
= P(x-y) + P(y-z) + P(z-x).
\end{array}
$$
$$
\begin{array}{l}
\text { 6. Let } Q(x, y, z) \\
=x P(x)+y P(y)+z P(z)- \\
\quad x y z(P(x-y)+P(y-z)+P(z-x)) .
\end{array}
$$

Then the condition indicates that for all real numbers \( x, y, z \) satisfying
$$
2 x y z=x+y+z \text { and } x y z \neq 0
$$

we have
$$
Q(x, y, z)=0 \text {. }
$$

Therefore, for all real numbers \( x, y \) satisfying
$$
x y \neq 0, x y \neq \frac{1}{2}, x+y \neq 0
$$

we have
$$
R(x, y)=Q\left(x, y, \frac{x+y}{2 x y-1}\right)=0 .
$$

This indicates that for any fixed \( y \), \( R(x, y) \) as a rational function of \( x \) has infinitely many zeros, hence \( R(x, y) \equiv 0 \) (where it is defined). Thus, if we consider the coefficients of the numerator and denominator of \( R(x, y) \) as complex numbers, we get that for all complex numbers \( x, y, z \) satisfying
$$
2 x y z=x+y+z \text { and } x y z \neq 0
$$

we have
$$
Q(x, y, z)=0 \text {. }
$$

Now, consider \( P \) and \( Q \) as polynomials with complex coefficients.
Taking \( (x, y, z)=(t, -t, 0) \), we get
$$
t P(t) - t P(-t) \equiv 0
$$
\(\Rightarrow P\) is an even function.
Taking \( (x, y, z)=\left(x, \frac{i}{\sqrt{2}}, -\frac{i}{\sqrt{2}}\right) \), we get
$$
\begin{array}{l}
x P(x) + \frac{i}{\sqrt{2}}\left(P\left(\frac{i}{\sqrt{2}}\right) - P\left(-\frac{i}{\sqrt{2}}\right)\right) \\
\equiv \frac{1}{2} x\left(P\left(x - \frac{i}{\sqrt{2}}\right) + P\left(x + \frac{i}{\sqrt{2}}\right) + P(\sqrt{2} i)\right) .
\end{array}
$$

Since \( P \) is an even function, the second term on the left side is zero.
Thus, \( P\left(x - \frac{i}{\sqrt{2}}\right) + P\left(x + \frac{i}{\sqrt{2}}\right) - 2 P(x) \equiv P(\sqrt{2} i) \).
Since the left side is the second-order difference of \( P \) and the right side is a constant, we have \( \operatorname{deg} P \leqslant 2 \).

Since \( \operatorname{deg} P \leqslant 2 \) and \( P \) is an even function, \( P(x) \) must be of the form \( P(x) = c x^2 + d \). Substituting into the condition, we find that the set of all \( P(x) \) satisfying the condition is
$$
P(x) = c(x^2 + 3) \text{ (where \( c \) is any real number)}.
$$