2. Find the smallest constant $C$, such that for all real numbers $x, y, z$ satisfying $x+y+z=-1$, we have
$$
\left|x^{3}+y^{3}+z^{3}+1\right| \leqslant C\left|x^{5}+y^{5}+z^{5}+1\right| \text {. }
$$
2. When $x=y=z=-\frac{1}{3}$, we have $C \geqslant \frac{9}{10}$.

Below is the proof: For all real numbers $x, y, z$ satisfying $x+y+z=-1$, we have
$$
\left|x^{3}+y^{3}+z^{3}+1\right| \leqslant \frac{9}{10}\left|x^{5}+y^{5}+z^{5}+1\right| \text {. }
$$

In fact,
Equation (1)
$$
\begin{array}{l}
\Leftrightarrow\left|x^{3}+y^{3}+z^{3}-(x+y+z)^{3}\right| \\
\leqslant \frac{9}{10}\left|x^{5}+y^{5}+z^{5}-(x+y+z)^{5}\right| \\
\Leftrightarrow 3|(x+y)(y+z)(z+x)| \\
\leqslant \frac{9}{10} \times 5|(x+y)(y+z)(z+x)| \text {. } \\
\left(x^{2}+y^{2}+z^{2}+xy+yz+zx\right) 1 . \\
\end{array}
$$

Since $z=-1-x-y$, we have
$$
\begin{array}{l} 
x^{2}+y^{2}+z^{2}+xy+yz+zx \\
= x^{2}+y^{2}+(-1-x-y)^{2}+xy+ \\
y(-1-x-y)+x(-1-x-y) \\
= x^{2}+xy+y^{2}+x+y+1 \\
=\left(x+\frac{y}{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}\left(y+\frac{1}{3}\right)^{2}+\frac{2}{3} \\
\geqslant \frac{2}{3} .
\end{array}
$$

Then the right side of Equation (2)
$$
\begin{array}{l}
\geqslant \frac{9}{10} \times 5|(x+y)(y+z)(z+x)| \frac{2}{3} \\
=3|(x+y)(y+z)(z+x)|=\text { left side. }
\end{array}
$$

Thus, Equation (2) holds.
Therefore, Equation (1) holds.
In summary, the smallest constant $C$ is $\frac{9}{10}$.