[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.
