find the domain and inverse for
(f)= 2 + 1/(e^(x/3-5)
the domain is all reals.
e^z is never zero, so we don't have to watch out for that.
Unless your typo conceals the fact that you means
f(x) = 2 + 1/(e^(x/3)-5)
In which case we cannot have
e^(x/3)-5 = 0
e^(x/3) = 5
x/3 = ln 5
x = 3 ln 5 = ln 125
For the inverse, swap variables and solve for y.
x = 2+1/(e^(y/3)-5)
x-2 = 1/(e^(y/3)-5)
e^(y/3)-5 = 1/(x-2)
e^(y/3) = 5 + 1/(x-2) = (5x-9)/(x-2)
y/3 = ln (5x-9)/(x-2)
y = 3 ln (5x-9)/(x-2)
Note that the domain of f^-1 excludes x=2, just as the range of f(x) does not include the value y=2 (an asymptote)