Note that the Product Rule and Quotient Rule imply that #f'(x)=3x^{2}e^{-6(x-3)}-6x^{3}e^{-6(x-3)}#.

Also note, by "educated guessing" (or looking at the graph), that #f(3)=3^{3}e^{-6(3-3)}=27*1=27#. Therefore, #f^{-1}(27)=3#. (These problems are typically set up in such a way that you will be "lucky" in your educated guessing on this part).

Since #f'(3)=3*3^{2}*1-6*3^{3}*1=27-162=-135#, it follows that #(f^{-1})'(27)=1/(f'(f^{-1}(27)))=- 1/135#.

The fact that #(f^{-1})'(y)=1/(f'(f^{-1}(y))# follows from the Chain Rule by differentiating both sides of the equation #f(f^{-1}(y))=y# if we assume the differentiability of the functions in question.

We are also implicitly assuming that #f# is invertible here, which in may only be locally so. In fact, this particular function is not invertible overall, since it fails the horizontal line test. However, near #x=3# it is invertible (the function has negative slope for #x>1/2#). See the graph below.

graph{x^3*e^(-6(x-3)) [-1, 7, -499973, 500027]}