\(f(x)=xe^{-x^2}\)
\(f'(x)=(1-2x^2)e^{-x^2}\)
\(f''(x)=(4x^3-6x)e^{-x^2}\)
\(f'(x)=0\) whenever \(1-2x^2=0\), which happens when \(x=\pm\dfrac1{\sqrt2}\). Only the positive root falls in the interval \([0,2]\). At this point, you have \(f''\left(\dfrac1{\sqrt2}\right)<0\) (concave downward). This indicates a maximum at \(\left(\dfrac1{\sqrt2},\dfrac1{\sqrt{2e}}\right)\approx0.7071,0.4289)\).
Meanwhile, \(f(0)=0\) and \(f(2)=2e^{-4}\approx0.0366\), so the maximum above must be absolute, and the absolute minimum occurs at the origin.
