9. Given the function $f(x)=\left(a+\frac{1}{a}\right) \ln x+\frac{1}{x}-x$.
(1) Suppose $a>1$, discuss the monotonicity of $f(x)$ on the interval $(0,1)$;
(2) Suppose $a>0$, find the extremum of $f(x)$.
(1) $f^{\prime}(x)=\left(a+\frac{1}{a}\right) \frac{1}{x}-\frac{1}{x^{2}}-1=-\left[\frac{1}{x^{2}}-\left(a+\frac{1}{a}\right) \frac{1}{x}+1\right]=-\left(\frac{1}{x}-a\right)\left(\frac{1}{x}-\frac{1}{a}\right)$.

Since $\frac{1}{x}-\frac{1}{a}>0$, $f(x)$ is monotonically decreasing on $\left(0, \frac{1}{a}\right)$ and monotonically increasing on $\left(\frac{1}{a}, 1\right)$.
(2) $f^{\prime}(x)=-\left(\frac{1}{x}-a\right)\left(\frac{1}{x}-\frac{1}{a}\right)$, when $0<1$ and $a>1$, $f(x)$ is monotonically decreasing on $\left(0, \frac{1}{a}\right)$, monotonically increasing on $\left(\frac{1}{a}, a\right)$, and monotonically decreasing on $(a,+\infty)$,
so $f(x)_{\text {min}}=f\left(\frac{1}{a}\right)=-\left(a+\frac{1}{a}\right) \ln a+a-\frac{1}{a}, f(x)_{\text {max}}=f(a)=\left(a+\frac{1}{a}\right) \ln a+\frac{1}{a}-a$.