Your question is not in good English so it is hard to know how to answer.
[MATH]y = \left ( \dfrac{(12^3m^{33}}{4^{-3}} \right )^2.[/MATH]
is a valid equation, but it can be expresed in ways that may be easier to understand. This is called "simplification" in English. There is no objective meaning to the term. It depends partially on what you need to do later and partly on your personal feeling about what is easy to understand.
In this case, you can get rid of the parentheses by doing the exponentiation as follows:
[MATH]y = \left ( \dfrac{12^3m^{33}}{4^{-3}} \right )^2 = \dfrac{12^{2*3}m^{2*33}}{4^{2*(-3)}} = \dfrac{12^6m^{66}}{4^{-6}}.[/MATH]
That still looks rather formidable to me. We can get rid of the fraction by eliminating the negative exponent:
[MATH]y = \dfrac{12^6m^{66}}{4^{-6}} = \dfrac{\dfrac{12^6m^{66}}{1}}{\dfrac{1}{4^6}} = \dfrac{12^6m^{66}}{1} * \dfrac{4^6}{1} = 4^6 * 12^6m^{66}.[/MATH]
That looks simpler to me than what we started with. Depending on what we want to do, there are now two ways to go. One way is to combine the product of numbers with like exponents:
[MATH]y = 4^6 * 12^6m^{66} = (4 * 12)^6m^{66} = 48^6m^{66}.[/MATH]
Does that look simpler to you than what we started with? It does to me.
However, if we know that all the numbers are integers, it might be more convenient for future work to simplify to the smallest integers possible. In that case we might go:
[MATH]y = 4^6 * 12^6m^{66} = (2^2)^6 * (2 * 2 * 3)^6m^{66} = \\
2^{12} * 2^6 * 2^6 * 3^6m^{66} = 2^{24} * 3^6m^{66}.[/MATH]That involves more terms than our previous answer, but two and three are easier numbers to imagine than forty-eight. (In fact, I cannot "see" in my mind 48 of anything, but I can "see" the difference in my mind between two cats and three cats.)
To summarize, all the expressions that we got for y are correct in the sense that they are valid arithmetically, but only two are as easy to understand as possible.