9. Given the function $f(x)=\left(\frac{x-1}{x+1}\right)^{2}(x>1)$, if the inequality $(1-\sqrt{x}) f^{-1}(x)>m(m-\sqrt{x})$ holds for every $x$ value in the interval $\left[\frac{1}{16}, \frac{1}{4}\right]$, find the range of real number $m$.
9. Solution: Given $f(x)=\frac{(x-1)^{2}}{(x+1)^{2}} \Rightarrow f^{-1}(x)=\frac{1+\sqrt{x}}{1-\sqrt{x}}$, since $f(x)=\left(\frac{x-1}{x+1}\right)^{2}=\left(1-\frac{2}{x+1}\right)^{2}$, and $x>1$, hence the range of $f(x)$ is $(0,1)$,
thus $f^{-1}(x)=\frac{1+\sqrt{x}}{1-\sqrt{x}}(0m(m-\sqrt{x})$,
hence $(1+m) \sqrt{x}+1-m^{2}>0$, for all $x$ satisfying $\frac{1}{16} \leqslant x \leqslant \frac{1}{4}$.
Clearly, $1+m \neq 0$, i.e., $m \neq-1$.
Let $t=\sqrt{x}$, then $g(t)=(1+m) t+1-m^{2}\left(\frac{1}{4} \leqslant t \leqslant \frac{1}{2}\right)$,
so $g(t)$ is a linear function, and $g(t)=(1+m) t+\left(1-m^{2}\right)>0$ for $t \in\left[\frac{1}{4}, \frac{1}{2}\right]$.
(1) When $m>-1$, it is sufficient that $g\left(\frac{1}{4}\right)>0 \Rightarrow-10 \Rightarrow m \in \varnothing$.

In summary, the range of $m$ is $-1<m<\frac{5}{4}$.