Hello Nyeirah. Here is how I read your problem:

((x^-3y^4z^-2)^2)/y^0z^-2 = ((x^-3y⁴z^-2)²)/y⁰z^-2

If that is a correct restatment of the problem, then the following is how I would solve it. (NOTE: my answer might differ from other tutor's answers because we read the problem you wrote differently. I am answering what I think you are asking.)

First, we need to get rid of some of those parentheses in the numerator. Recall that (A^m)ⁿ =A^mn .

((x^-3y⁴z^-2)²)/y⁰z^-2

= x^-6y⁸z^-4/y⁰z^-2

Now for the denomiator, y⁰ is just 1. Anything to the zero power is just 1.

=x^-6y⁸z^-4/z^-2

When the only operations between terms are multiplication and division, like we have here, then you can move terms from denominator to numerator by changing the sign of the exponent. We will move the z^-2 term to the numerator and change its exponent from -2 to 2.

=x^-6y⁸z^-4z²

Now combine those two z terms. The rule is A^mAⁿ=A^m+n. Here z^-4z² = z^-4+2 = z^-2

x^-6y⁸z^-2 [Answer]

We can rearrange if we like to put all the negative exponent terms in the denominator.

y⁸/x⁶z² [Answer, in a different form]

Let's look at this in two parts: Numerator and Denominator

Demoninator first b/c its easy

0*z=0

0^-2=0

Therefore y⁰=1 because any number taken to the power 0 is equal to 1

So your denominator is 1

Numerator:

Let's start inside the parenthese to see what we can simplify.  Starting at the highest exponent 4z^-2

4z^-2=1/(4z²)

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

so you are left with (1/(x^3y^(1/4z²)))²=(1/(x^6y^(1/4z²)))  since when you square an exponential function you mutiply the square (or 2) by the exponent