Answer
$\frac{25x^{12}}{81y^{6}}, y\ne0$
Work Step by Step
Based on the power of a quotient rule, we know that $(\frac{a}{c})^{n}=\frac{a^{n}}{c^{n}}$ (where $n$ is a positive integer and $a$ and $c$ are real numbers).
Therefore, $(\frac{5x^{6}}{9y^{3}})^{2}=\frac{5^{2}\times (x^{6})^{2}}{9^{2}\times (y^{3})^{2}}$.
Based on the power of a product rule, we know that $(ab)^{n}=a^{n}b^{n}$ (where $n$ is a positive integer and $a$ and $b$ are real numbers).
Therefore, $\frac{5^{2}\times (x^{6})^{2}}{9^{2}\times (y^{3})^{2}}=\frac{25x^{6\times2}}{81\times y^{3\times2}}=\frac{25x^{12}}{81y^{6}}, y\ne0$.
