Let $x, y, z \in(0,+\infty)$, and $x^{2}+y^{2}+$ $z^{2}=1$. Find the range of the function $x+y+z-x y z$
First, prove $f>1$: Clearly, $0<y^{2}, z>z^{2}$. Thus,
$$\begin{aligned}
f & =x+y+z-x y z \\
& \geqslant x+y+z-x \cdot \frac{y^{2}+z^{2}}{2} \\
& =x+y+z-x \cdot \frac{1-x^{2}}{2}\left(\text { because } x^{2}+y^{2}+z^{2}\right. \\
& =1) \\
& =x \cdot \frac{1+x^{2}}{2}+y+z \\
& \geqslant x \cdot x+y+z \\
& >x^{2}+y^{2}+z^{2}=1
\end{aligned}$$

Therefore, $f>1$, and it is easy to see that when $x \rightarrow 1$ or $y \rightarrow 1$ or $z \rightarrow 1$ (satisfying $x^{2}+y^{2}+z^{2}=1$), $f \rightarrow 1$.

Next, prove $f \leqslant \frac{8 \sqrt{3}}{9}$: Given that positive numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$ and by the pigeonhole principle, at least two of $x, y, z$ are not less than (or not greater than) $\frac{1}{\sqrt{3}}$. Without loss of generality, assume they are $x, y$. We have
$$\left(x-\frac{1}{\sqrt{3}}\right)\left(y-\frac{1}{\sqrt{3}}\right) \geqslant 0$$

This can be rewritten as
$$\frac{1}{\sqrt{3}}(x+y) \leqslant \frac{1}{3}+x y \text {, i.e., } x+y-\sqrt{3} x y \leqslant \frac{1}{\sqrt{3}}$$

From this, we get
$$\begin{array}{l}
y z+z x+x y-\sqrt{3} x y z \\
=x y+(x+y-\sqrt{3} x y) z \\
\leqslant x y+\frac{1}{\sqrt{3}} z \leqslant \frac{x^{2}+y^{2}}{2}+\frac{\frac{1}{3}+z^{2}}{2} \\
=\frac{1}{2}\left(x^{2}+y^{2}+z^{2}+\frac{1}{3}\right)=\frac{2}{3}\left(\text { because } x^{2}+y^{2}+\right.
\end{array}$$
$$\left.z^{2}=1\right)$$

Also,
$$\begin{array}{l}
2 \cdot \sqrt{3}(x+y+z) \leqslant(\sqrt{3})^{2}+(x+y+z)^{2} \\
=3+\left(x^{2}+y^{2}+z^{2}\right)+2(y z+z x+x y) \\
=3+1+2(y z+z x+x y)
\end{array}$$

That is,
$$x+y+z \leqslant \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}(y z+z x+x y)$$

Combining (2) and (3), we have
$$\begin{array}{l}
x+y+z-x y z \\
\leqslant \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}(y z+z x+x y-\sqrt{3} x y z) \\
\leqslant \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}} \cdot \frac{2}{3}=\frac{8}{3 \sqrt{3}}
\end{array}$$

That is, $f \leqslant \frac{8 \sqrt{3}}{9}$, and the equality holds if and only if $x=y=z=\frac{1}{\sqrt{3}}$.

In summary, the range of $f$ is $\left[1, \frac{8 \sqrt{3}}{9}\right]$.