15. The three lines $y=\sqrt{2} x, y=-\sqrt{2} x$ and $x=m$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into several parts. Now, using 6 different colors to color these parts, each part is colored with one color, and any two parts are of different colors, there are 720 different coloring methods. Then the range of the real number $m$ is $\qquad$
15. $|m|-2<m \leqslant-\frac{2}{3}$ or $m=0$ or $\left.\frac{2}{3} \leqslant m<2\right\}$.

Obviously, the two fixed lines $y= \pm \sqrt{2} x$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into 4
blocks (as shown in Figure 5). When $|m|$
$\geqslant 2$, using 6 colors to color these 4 blocks, the number of different coloring methods is $\mathrm{A}_{6}^{4}<$
720, which does not meet the requirement, so, $|m|<2$.
$$
\text { From }\left\{\begin{array}{l}
y= \pm \sqrt{2} x, \\
\frac{x^{2}}{4}+y^{2}=1
\end{array} \Rightarrow x= \pm \frac{2}{3}\right. \text {. }
$$

When $m=0$, the line $x=0$ passes through the origin, at this time, the elliptical region is divided into 6 blocks, using 6 colors to color these 6 blocks, the number of different coloring methods is $\mathrm{A}_{6}^{6}=720$, which meets the requirement.

When $0<|m|<\frac{2}{3}$, the line $x=m$ divides 3 of the blocks into two each, resulting in 7 blocks. Using 6 colors to color these 7 blocks, there must be two blocks of the same color, which does not meet the requirement.

When $\frac{2}{3} \leqslant|m|<2$, the line $x=m$ divides one of the blocks into two, resulting in 5 blocks. Using 6 colors to color these 5 blocks, the number of different coloring methods is $\mathrm{A}_{6}^{5}=720$, which meets the requirement.
In summary, the range of values for $m$ is
$$
\left\{m \left\lvert\,-2<m \leqslant-\frac{2}{3}\right. \text { or } m=0 \text { or } \frac{2}{3} \leqslant m<2\right\} \text {. }
$$