= Solution
The <derivative> is $f'(x)=2x\cos(x^2-2)$. Thus a <stationary point> occurs either at $x=0$ or where $x^2-2=\pi/2+k\pi$. Since $-2\leq x^2-2<2$ on the specified open interval, the only allowed cosine zeros are $-\pi/2$ and $\pi/2$. The five stationary abscissae are
$$
\boxed{x=0,\qquad x=\pm\sqrt{2-\frac\pi2},\qquad
x=\pm\sqrt{2+\frac\pi2}}.
$$
Their ordinates are $-\sin2$, $-1$, and $1$, respectively. For completeness, $f''=2\cos(x^2-2)-4x^2\sin(x^2-2)$ shows that the inner pair are <local minima> and the outer pair are <local maxima>; $x=0$ is also a <local maximum> because $2\cos2<0$.
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