Suppose #a# to be a positive value except for #1#.

If #a# is larger than #1#, the following is true.

i) If #1 < a#, #a^x < a^y# if and only if #x < y#.

However, if #a# is smaller than 1, the magnitude reverces.

ii) If #0 < a < 1#, #a^x < a^y# if and only if #x > y#.

Therefore, you must solve the inequation as follows:

#1)# When #x# satisfies #0<(x^2-x+1) < 1#

You have already solved #0<(x^2-x+1) < 1# and got #0< x<1#.

This time,

#(x^2-x+1)^-1 ≦ (x^2-x+1)^x # ⇔ #-1 ≧ x#

The common part of #0< x < 1# and #-1 ≧ x# is null.

#2)# When #x# satisfies #(x^2-x+1)=1 #, #x=0,1#.

Since #1^n# is always 1, this satisfies the inequation.

#3)# When #x# satisfies #1<(x^2-x+1) #, the range of #x# is

#x< 0# or #1< x#.

At that time,

#(x^2-x+1)^-1 ≦ (x^2-x+1)^x # ⇔ #-1 ≦ x#

and the common part of (#x< 0 # or #1< x#) and #-1 < x#

is #-1≦ x < 0# or #1< x#.

Now you got to the answer #-1≦ x ≦ 0# or #1≦ x#.

The answer was checked at http://www.wolframalpha.com