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

From the condition, we know $x \in[-1,1]$.
Let $x=\cos \alpha(\alpha \in[0, \pi])$. Then
$$
\begin{aligned}
y & =\frac{\sin \alpha}{2+\cos \alpha}(y \geqslant 0) \\
\Rightarrow & 2 y=\sin \alpha-y \cos \alpha \\
& =\sqrt{1+y^{2}} \sin (\alpha+\theta) \\
& \leqslant \sqrt{1+y^{2}}
\end{aligned}
$$
$$
\Rightarrow 1+y^{2} \geqslant 4 y^{2} \Rightarrow y^{2} \leqslant \frac{1}{3} \text {. }
$$

Since $y \geqslant 0$, therefore, $y \in\left[0, \frac{\sqrt{3}}{3}\right]$.