1. The range of the function $f(x)=\frac{\left(x-x^{3}\right)\left(1-6 x^{2}+x^{4}\right)}{\left(1+x^{2}\right)^{4}}$ is . $\qquad$
,- 1. $\left[-\frac{1}{8}, \frac{1}{8}\right]$.
Since the domain of the function $f(x)$ is $\mathbf{R}$, we let
$$
x=\tan \frac{\alpha}{2}(\alpha \in \mathbf{R}) \text {. }
$$

Then $\sin \alpha=\frac{2 x}{1+x^{2}}, \cos \alpha=\frac{1-x^{2}}{1+x^{2}}$.
Given $f(x)=\frac{\left(x-x^{3}\right)\left(x^{4}-6 x^{2}+1\right)}{\left(1+x^{2}\right)^{4}}$
$=\frac{2 x}{1+x^{2}} \cdot \frac{1-x^{2}}{1+x^{2}}\left(\left(\frac{1-x^{2}}{1+x^{2}}\right)^{2}-\frac{1}{2}\right)$
$=\frac{1}{2} \sin \alpha \cdot \cos \alpha\left(2 \cos ^{2} \alpha-1\right)$
$=\frac{1}{4} \sin 2 \alpha \cdot \cos 2 \alpha=\frac{1}{8} \sin 4 \alpha$
$$
\Rightarrow f(x) \in\left[-\frac{1}{8}, \frac{1}{8}\right] .
$$