3. The range of the function $f(x)=\frac{\sqrt{1-x^{2}}}{x+2}$ is
3. $\left[0, \frac{\sqrt{3}}{3}\right]$.

Let $y=\sqrt{1-x^{2}}(-1 \leqslant x \leqslant 1)$. Then the image of point $P(x, y)$ is the part of the unit circle above the $x$-axis (as shown in Figure 3).
Rewrite the function as
$$
\begin{array}{l}
f(x) \\
=\frac{y-0}{x-(-2)},
\end{array}
$$

which represents the slope of the line $AP$ connecting the fixed point $A(-2,0)$ and the point $P(x, y)$ on the semicircle.
It is easy to see that, in the right triangle $\triangle AOP$,
$$
OP=1, OA=2, \angle OAP=30^{\circ} \text{. }
$$

Therefore, $k_{AP}=\frac{\sqrt{3}}{3}, k_{AO}=0$.
Hence, $0 \leqslant f(x) \leqslant \frac{\sqrt{3}}{3}$.