∴[f(- 1)e^(- 1)]'=e^(- 1)[(f'(- 1)+f(- 1)]=0
Then f'(- 1)+f(- 1)=0.
In A and B, f'(- 1)=f(- 1)=0, which is consistent.
In c language, f'(- 1)>0, f (-1) < 0, f'(- 1)+f(- 1)=0 is possible.
In d, f'(- 1)>0, f(- 1)>0, then f'(- 1)+f(- 1)>0, which does not match.
Therefore, choose D in this question.