​ Help asap, please!

$\frac{\frac{a^2}{3^2}}{\frac{3^2-a^2}{3a^2}}=\frac{3a^2a^2}{3^23^2-3^2a^2}=\frac{a^4}{3^3-3a^2}$

Lets work with this, first

3                  1

____    -     ___

a^2             3

Cross-multiply ans keep the same sign as between the fractions

3*3 - 1*a^2   =  9 - a^2

Take the product of the denominators 

3a^2

So   we have

( 9 - a^2) 

________

    3a^2

So...puitting this together, we have

[ a^2 / 9 ]  ÷  [ ( 9 - a^2) / ( 3a^2) ]

Invert the second fraction and multiply

a^2               3a^2                      3a^4

___   *         ______    =           _______   

9                  9 - a^2                  9(9 - a^2)

Cancel  the 3 in the numerator and the 9 in the denominator  and we are left with

a^4                        a^4

_______     =      ________

3(9 - a^2)           27 - 3a^2