There's three ways it can be done:

1) Use the definition of the Taylor series at #x=a#:

#f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots#.

In the present example, #f'(x)=1-3x^2#, #f''(x)=-6x#, #f'''(x)=-6#, and all higher-order derivatives are zero. Since #f(-3)=24#, #f'(-3)=-26#, #f''(-3)=18#, #f'''(-3)=-6#, and all higher derivatives are zero, the answer is:

#24-26(x+3)+9(x+3)^2-(x+3)^3#, and this does indeed equal #f(x)=x-x^3# for all #x#.

2) You could do a change-of-variables that has the effect of translating the origin to the left by 3 units: #u=x+3# so that #x=u-3#. The coefficients of #f(u-3)# as a Taylor series centered at #u=0# would be the same coefficients of #f(x)# as a Taylor series centered at #x=-3#. Here's the expansion calculation:

#f(u-3)=(u-3)-(u-3)^3=u-3-(u^3-9u^2+27u-27)=24-26u+9u^2-u^3#

Hence, #f(x)=24-26(x+3)+9(x+3)^2-(x+3)^3#.

3) You could realize that since #f(x)=x-x^3# is cubic, its Taylor series about any point will be cubic as well. In particular, about #x=-3# it will be of the form #a+b(x+3)+c(x+3)^2+d(x+3)^3#. You can then expand this out:

#a+bx+3b+cx^2+6cx+9c+dx^3+9dx^2+27dx+27d=
(a+3b+9c+27d)+(b+6c+27d)x+(c+9d)x^2+dx^3#

Set these coefficients equal to the corresponding coefficients of #x-x^3# to get a system of 4 equations and 4 unknowns: #d=-1#, #c+9d=0#, #b+6c+27d=1#, and #a+3b+9c+27d=0#. This system is already in "triangular form" (if your form the corresponding matrix ) and is easily solved by "back substitution": #d=-1#, #c=-9d=9#, #b=1-6c-27d=1-54+27=-26#, and #a=-3b-9c-27d=78-81+27=24# to get the same answer: #24-26(x+3)+9(x+3)^2-(x+3)^3#.