 Solution

For $f(x)=x^3{e}^{-3x^2/2}/3$, the product rule gives $f^{\prime }(x)=x^2(1-x^2)e^{-3x^2/2}$, so the stationary points occur at $x=-1,0,1$. Its sign is negative for $|x|>1$ and positive for $0<|x|<1$. ConsequentlyThey 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 $f(x)=x^3/3+O(x^5)$, $(0,0)$ is a stationary inflection point, not an extremum.