125^(5x-9y-55)=(5^3(5x-9y-55))=(5²)(5^(15x-27y-167)).
This is because (5²)(5^(15x-27y-167))=(5^(15x-27y-167+2))=5^(15x-27y-165).
So we can factorise:
f(x,y)=1/(5²(1-5^(15x-27y-167))).
5^(15x-27y-167)=1 when 15x-27y-167=0. This would give us a denominator of 5²(1-1)=0 so 15x-27y-167≠0 is the domain of f(x,y). The function can be evaluated (exists) for all x,y apart from 15x-27y-167=0.