f(x)=18xe^(1-x^2), so f'(x)=18e^(1-x^2)+18xe^(1-x^2)*-2x=18(e^(1-x^2))(1-2x^2).
When f'(x)=0 we have a stationary point, so, since the exponential cannot be zero, 2x^2=1 and x=+sqrt(2)/2.
f(-sqrt(2)/2)=-9sqrt(2)e^(1/2)=-20.985 and f(sqrt(2)/2)=20.985. Therefore the extrema are (-0.7071,-20.985) and (0.7071,20.985). The exact values are (-sqrt(2)/2,-9sqrt(2e)) and (sqrt(2)/2,9sqrt(2e)). The stationary point appears to be a point of inflection in each case, since the values of f(x) do not change signs near the points. The absolute extrema are the same as the local extrema and have the same magnitude.