16. (9th "Hope Cup" Invitational Training Question) If the function $f\left(x-\frac{1}{x}\right)=\frac{x}{x^{2}-1}-x^{2}-\frac{1}{x^{2}}(x \neq 0$, $x \neq \pm 1)$, then $f(x)=$ $\qquad$ .
16. Fill in $-x^{2}+\frac{1}{x}-2$. Reason: From $f\left(x-\frac{1}{x}\right)=\frac{x}{x^{2}-1}-x^{2}-\frac{1}{x^{2}}=\frac{1}{x-\frac{1}{x}}-\left[\left(x-\frac{1}{x}\right)^{2}+2\right]=$ $-\left(x-\frac{1}{x}\right)^{2}-\left(x-\frac{1}{x}\right)^{-1}-2$. Therefore, $f(x)=-x^{2}+\frac{1}{x}-2$.