Does there exist a real number $k$, such that
$$
f(x, y, z)=\frac{x^{5}+y^{5}+z^{5}+k\left(x^{2}+y^{2}+z^{2}\right)\left(x^{3}+y^{3}+z^{3}\right)}{x+y+z}
$$

is a polynomial in three variables.
Assume there exists such a real number $k$. Then
$$
x^{5}+y^{5}+z^{5}+k\left(x^{2}+y^{2}+z^{2}\right)\left(x^{3}+y^{3}+z^{3}\right)
$$

has a factor $x+y+z$.
Let $y=z=1$. Then $x+2$ is a factor of
$$
x^{5}+2+k\left(x^{3}+2\right)\left(x^{2}+2\right)
$$

Perform polynomial division.
The remainder of dividing $(1+k) x^{5}+2 k x^{3}+2 k x^{2}+4 k+2$ by $x+2$ is $-36 k-30$. For $x+2$ to divide $x^{5}+2+k\left(x^{3}+2\right)\left(x^{2}+2\right)$, we have
$$
-36 k-30=0 \Rightarrow k=-\frac{5}{6} \text {. }
$$

Therefore, $f(x, y, z)$
$$
\begin{array}{l}
= \frac{x^{5}+y^{5}+z^{5}-\frac{5}{6}\left(x^{2}+y^{2}+z^{2}\right)\left(x^{3}+y^{3}+z^{3}\right)}{x+y+z} \\
= \frac{\frac{1}{6} x^{5}-\frac{5}{6}\left(y^{2}+z^{2}\right) x^{3}-\frac{5}{6}\left(y^{3}+z^{3}\right) x^{2}+\frac{1}{6}\left(y^{5}+z^{5}-5 y^{3} z^{2}-5 y^{2} z^{3}\right)}{x+y+z} \\
= \frac{1}{6} x^{4}-\frac{1}{6}(y+z) x^{3}-\frac{1}{3}\left(2 y^{2}+2 z^{2}-y z\right) x^{2}- \\
\frac{1}{6}(y+z)\left(y^{2}-3 y z+z^{2}\right) x+ \\
\frac{1}{6}(y+z)^{2}\left(y^{2}-3 y z+z^{2}\right) .
\end{array}
$$

In conclusion, such a real number $k$ exists, and $k=-\frac{5}{6}$.