Simply this expression?

Note that your equation above and in the link are different (the -2 power goes to the top and bottom in the JPG, but only the top in this link).

Assuming the JPG is correct, first simplify the part in the parenthis but subtracting like exponents or....

[3x^(3/2 - 2) y^(3 - (-0.5))]^-2

[3x^(-0.5) y^(3.5)]^-2

Now, multiply the exponents by -2

[3^(-2) x^(1) y^(-7)]

This can be farther simplified, as x^1 = x, and y^(-7) can be brought down to the denominator (as can the 3^-2), so its simplist form is ...

x / (9 y^7)

~

Note, the other two answers are wrong. Rubiac because he forgot to apply the -2 power to the 3 in the top, and Harpy because he messed up subtracting the negative exponent for the y.

[3(x^3/2)y^3]^-2
_________
[x^2(y^-1/2)]

[3(x^3/2)y^3(y^1/2)] ^ - 2
_________________
[x^2]

[3(x^2)(x^1/2)y^3(y^1/2)] ^ - 2
_____________________ --> x^2 will be
[x^2]

[3(x^1/2)y^3(y^1/2)] ^ - 2 --> since the exponential is negative

1
_________________ ---> since (a^(1/n))^n= a
[3(x^1/2)y^3(y^1/2)] ^ 2



1
_______ --> this is the simplified form
[9(x)y^3]

((3(x^3/2)y^3)/(x^2(y^-1/2)))^-2=

((x^2(y^-1/2))/(3(x^3/2)y^3))^2=

((x^(2-3/2)(y^(-1/2-3))/3)^2=

((x^1/2)(y^-7/2)/3)^2=

((x^1/2)/9(y^7/2))^2=

x/9(y^7)

Note
The question you have typed IS NOT THE SAME as the question on link.

I will go with link version :-

[ 3x^(3/2) y³ ] ^(- 2)
------------------------
[ x² y^(-1/2) ] ^(- 2)

[ 3 y^(7/2) ] ^(- 2)
---------------------
[ x^(1/2) ] ^(-2)

x
-------
9 y^7

[3x^(3/2)y^3]^-2/[x^2y^(-1/2)]
= [3^-2x^(3/2 * -2)y^(3 *- 2)]/[x^2y^(-1/2)]
= [3^(-2)x^(-3)y^(-6)]/[x^2y^(-1/2)]
= [3^(-2)/1][x^(-3)/x^2][y^(-6)/y^(-1/2)]
= [1/(3^2)][x^(-3 - 2)][y^(-6 + 1/2)]
= [1/9][x^(-5)][y^(-11/2)]
= [1/9][1/x^5][y^(-11/2)]
= [1/9x^5][√(y^-11)]
= [1/9x^5][√(1/y^11)]
= [1/9x^5][√(1/y^5 * 1/y^5 * 1/y)]
= [1/9x^5][1/y^5√(1/y)]
= √(1/y)/9x^5y^5