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