= Solution
For $f(x)=x^3e^{-3x^2/2}/3$, the <product rule> gives $f'(x)=x^2(1-x^2)e^{-3x^2/2}$, so the <stationary points> occur at $x=-1,0,1$. Its sign is negative for $|x|>1$ and positive for $0<|x|<1$. Consequently
$$
\boxed{\left(-1,-\frac{e^{-3/2}}3\right)\text{ is a minimum},\qquad\left(1,\frac{e^{-3/2}}3\right)\text{ is a maximum}.}
$$
They are also the global <extrema>, since the function tends to zero at both infinities and is a <monotone function> on the intervals just identified. At zero the <derivative> does not change sign. Since $f(x)=x^3/3+O(x^5)$, \b[$(0,0)$ is a stationary <inflection point>], not an <extremum>.
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