Real numbers $x, y$ satisfy $x^{2}+x y+y^{2}=1, F(x, y)=x^{3} y+x y^{3}$.
(1) Prove: For any $x, y \in \mathbf{R}$ that meet the requirements, $F(x, y) \geqslant-2$.
(2) Find the maximum possible value of $F(x, y)$.
15. (1) From the condition, we have $(x+y)^{2}=1+x y \geqslant 0$, so $x y \geqslant-1$; also $x^{2}+y^{2}=1-x y \geqslant 0$, so $x y \leqslant 1$, i.e., $x y \in[-1,1]$.

Notice that, $F(x, y)=x y\left(x^{2}+y^{2}\right)=z(1-z)$, where $z=x y \in[-1,1]$. Thus, $F(x, y)=$ $z-z^{2}=(z+1)(2-z)-2$, when $-1 \leqslant z \leqslant 1$, we have $z+1 \geqslant 0,2-z \geqslant 0$, hence $F(x, y) \geqslant-2$.
(2) Let the maximum value of $F(x, y)$ be $A$, notice that when $x=y= \pm \frac{\sqrt{3}}{3}$, $F(x, y)=\frac{2}{9}$, so $A \geqslant \frac{2}{9}$.
On the other hand, let $x, y \in \mathbf{R}$, and $x^{2}+x y+y^{2}=1$, such that $F(x, y)=A$, then $t_{1}(=x y), t_{2}\left(=x^{2}+y^{2}\right)$ are the two roots of the equation
$$
t^{2}-t+A=0
$$

Notice that $A \geqslant \frac{2}{9}$, so $t_{1}, t_{2}$ are both positive. And $t_{2}=x^{2}+y^{2} \geqslant 2 x y=2 t_{1}$, so $t_{2} \geqslant \frac{2}{3}, t_{1} \leqslant \frac{1}{3}$ (here we use $t_{1}+t_{2}=1$). This indicates that the two roots $t_{1}, t_{2}$ of the polynomial $g(t)=t^{2}-t+A$ satisfy $00$, so it should be that
$$
\left\{\begin{array}{l}
g\left(\frac{1}{3}\right) \leqslant 0, \\
g\left(\frac{2}{3}\right) \leqslant 0 .
\end{array}\right.
$$

This indicates that $A \leqslant \frac{2}{9}$.
In summary, the maximum value of $F(x, y)$ is $\frac{2}{9}$.