15. Let $P$ be a point on the ellipse $2 x^{2}+3 y^{2}=1$, $F_{1}$ and $F_{2}$ be the two foci of the ellipse, and $\angle F_{1} P F_{2}=\frac{\pi}{6}$. Then the area of $\triangle F_{1} P F_{2}$ is . $\qquad$
15. $\frac{2-\sqrt{3}}{3}$.

Let the lengths of the semi-major axis and semi-minor axis of the ellipse be $a$ and $b$, respectively. Then $a=\frac{\sqrt{2}}{2}, b=\frac{\sqrt{3}}{3} \Rightarrow c=\frac{\sqrt{6}}{6}$. Therefore, $\left|F_{1} F_{2}\right|=\frac{\sqrt{6}}{3},\left|P F_{1}\right|+\left|P F_{2}\right|=\sqrt{2}$.
Applying the cosine rule in $\triangle P F_{1} F_{2}$, we get
$$
\begin{array}{l}
\left(\frac{\sqrt{6}}{3}\right)^{2}=\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-2\left|P F_{1}\right|\left|P F_{2}\right| \cos \frac{\pi}{6} \\
\Rightarrow\left|P F_{1}\right|\left|P F_{2}\right|=\frac{4(2-\sqrt{3})}{3} .
\end{array}
$$

Thus, $S_{\triangle F_{1} P F_{2}}=\frac{1}{2}\left|P F_{1}\right|\left|P F_{2}\right| \sin \frac{\pi}{6}=\frac{2-\sqrt{3}}{3}$.