Let's deal with each factor in the numerator individually first: when evaluating the power of a power, you have to multiply the exponents: [tex](a^b)^c = a^{bc}[/tex]
So, we have
[tex] [tex](3^2)^{-\frac{1}{2}} = 3^{2\cdot(-\frac{1}{2})} = 3^{-1},\quad (9^4)^{-1} = 9^{-4}[/tex]
Now remember that [tex]9=3^2[/tex] and [tex]27=3^3[/tex] to rewrite the expression as
[tex]\dfrac{3^{-1}\cdot (3^2)^{-4}}{(3^3)^{-3}} = \dfrac{3^{-1}\cdot 3^{-8}}{3^{-9}}[/tex]
Now use the rules
[tex]a^b\cdot a^c = a^{b+c},\quad \dfrac{a^b}{a^c} = a^{b-c}[/tex]
to conclude
[tex]\dfrac{3^{-1}\cdot 3^{-8}}{3^{-9}} = 3^{-1-8-(-9) = 3^0 = 1[/tex]
