9 Given three sets, $A=\{x \mid x \in \mathbf{R}$ and $x>1\}, B=\left\{y \left\lvert\, y=\lg \frac{2}{x+1}\right., x \in \mathrm{A}\right\}$, $C=\left\{z \left\lvert\, z=\frac{m^{2} x-1}{m x+1}\right., x \in \mathrm{A}\right\}$. If $C \subseteq B$, then the range of values for $m$ is $\qquad$.
$9(-\infty,-1] \cup\{0\}$. It is easy to know that $B=\{y \mid y<0\}$. When $y>1$, it always holds that $\frac{m^{2} x-1}{m x+1}<0$. When $m>0$, it is obviously impossible to have $f(x)<0$ for all $x>1$. When $m<0$, for $f(x)<0$ to always hold for $x>1$, it must be that $\left\{\begin{array}{l}\frac{1}{m^{2}} \leqslant 1 \\ -\frac{1}{m} \leqslant 1\end{array}\right.$, solving which gives $m \leqslant-1$. Therefore, the range of values for $m$ is $(-\infty,-1] \cup\{0\}$.