If (r^2/4) (3x)^r (2/9x^2)^(6-r) can be simplified to k/x^3, find the values of the constants?

The answer is as follows:

I'm confused about one part:  2/9x^2.  Is that (2/9) * x^2, or is it 2/(9x^2)?

I'm presuming that r^2/4 is really (1/4) * r^2

(1/4) * r^2 * (3^r) * (x^r) * 2^(6 - r) / (9x^2)^(6 - r) =>
(1/4) * (r^2) * (3^r) * (2^6 / 2^r) * (1/9)^(6) * (1/9)^(-r) * x^(-2 * (6 - r)) =>
(1/4) * (r^2) * (3^r / 2^r) * (2^6 / 9^6) * 9^r * x^(r - 12) * x^r =>
2^(6 - 2) * 9^(-6) * 9^(r) * r^2 * (3/2)^r * x^(2r - 12)

x^(2r - 12) = x^(-3)
2r - 12 = -3
2r = 9
r = 4.5

2^(4) * 9^(-6) * 9^(4.5) * (4.5)^2 * (3/2)^(4.5) =>
2^(4) * 3^(-12) * 3^(9) * (9/2)^2 * 3^(4.5) * 2^(-4.5) =>
2^(4) * 3^(-12) * 3^(9) * 9^2 * 2^(-2) * 3^(4.5) * 2^(-4.5) =>
2^(4 - 2 - 4.5) * 3^(-12 + 9 + 4 + 4.5) =>
2^(-2.5) * 3^(5.5) =>
3^(11/2) / 2^(5/2) =>
(3^11 / 2^5)^(1/2)

If you mean (2/9) * x^2

(1/4) * r^2 * 3^(r) * x^(r) * (2/9)^(6 - r) * (x^(-2))^(6 - r) =>
(2^(-2)) * r^2 * (3^r) * 2^(6 - r) * 9^(r - 6) * x^(2r - 12) * x^r =>
2^(-2 + 6 - r) * r^2 * (3^r) * 3^(2r - 12) * x^(2r - 12 + r) =>
2^(4 - r) * r^2 * 3^(3r - 12) * x^(3r - 12)

x^(3r - 12) = x^(-3)
3r - 12 = -3
3r = 9
r = 3

2^(4 - 3) * 3^2 * 3^(3 * 3 - 12) =>
2^(1) * 3^2 * 3^(9 - 12) =>
2 * 3^(2 - 3) =>
2 * 3^(-1) =>
2/3