f'(x)=-2xe^x-x^2e^x=-xe^x(2-x)=0

x 1=0，x2=2

F'(x)>0 to obtain x>2, where x<0 indicates monotonically increasing intervals of (-infinity, 0) and (2,+infinity).

f '(x)& lt； 0 gets 0.

So there is a maximum at X=0, that is, f(0)=0.

And f (-3) =-(-3) 2e (-3) =-9/e 3, and f (1) =-e.

So the maximum value on [-3, 1] is f(0)=0, and the minimum value is f (1) =-e.