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[tex]f(x)=xe^{-x^2/128}[/tex]
[tex]\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}[/tex]

Extrema can occur when the derivative is zero or undefined.

[tex]\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}=0\implies 1-\dfrac1{64}x^2=0\implies x^2=64\implies x=\pm8[/tex]

Maxima occur where the first derivative is zero and the second derivative is negative; minima where the second derivative is positive. You have

[tex]f''(x)=-\dfrac1{32}xe^{-x^2/128}+\left(1-\dfrac1{64}x^2\right)\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(-\dfrac3{64}x+\dfrac1{4096}x^3\right)e^{-x^2/128}[/tex]

At the critical points, you get

[tex]f''(-8)=\dfrac1{4\sqrt e}>0[/tex]
[tex]f''(8)=-\dfrac1{4\sqrt e}<0[/tex]

So you have a minimum at [tex]\left(-8,-\dfrac8{\sqrt e}\right)[/tex] and a maximum at [tex]\left(8,\dfrac8{\sqrt e}\right)[/tex].

Meanwhile, as [tex]x\to\pm\infty[/tex], it's clear that [tex]f(x)\to0[/tex], so these extrema are absolute on the function's domain.

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