3. If the point $P$ on the curve $x^{2}-y^{2}=9$ is 2016 units away from the line $y=-x$, then the distance from point $P$ to the angle bisector of the first and third quadrants is $\qquad$
3. $\frac{1}{448}$.

Let point $P(m, n)$, the distance from $P$ to the line $y=-x$ is $\frac{|m+n|}{\sqrt{2}}$, and the distance from $P$ to the line $y=x$ is $\frac{|m-n|}{\sqrt{2}}$.
Then $\frac{|m+n|}{\sqrt{2}} \cdot \frac{|m-n|}{\sqrt{2}}=\frac{\left|m^{2}-n^{2}\right|}{2}=\frac{9}{2}$.
Also, the distance from point $P$ to the line $y=-x$ is 2016, thus, the required distance is
$$
\frac{9}{2} \times \frac{1}{2016}=\frac{1}{448} .
$$