Ii. Simplify (sqrt(9x ^ 4 y ^ 2))/(root(27x ^ 6 y ^ 12, 3))

The numerator is $\sqrt{9x^4y^2}$.
We can simplify each term under the square root:

$$\sqrt{9x^4y^2} = \sqrt{9} \cdot \sqrt{x^4} \cdot \sqrt{y^2}$$ $$= 3 \cdot x^{4/2} \cdot y^{2/2}$$ $$= 3x^2y$$

The denominator is $\sqrt[3]{27x^6y^{12}}$.
We can simplify each term under the cube root:

$$\sqrt[3]{27x^6y^{12}} = \sqrt[3]{27} \cdot \sqrt[3]{x^6} \cdot \sqrt[3]{y^{12}}$$ $$= 3 \cdot x^{6/3} \cdot y^{12/3}$$ $$= 3x^2y^4$$

Now, substitute the simplified expressions back into the fraction:

$$\frac{3x^2y}{3x^2y^4}$$

Cancel out the common terms 3 and $x^2$:

$$= \frac{y}{y^4}$$

Using the rule of exponents $\frac{a^m}{a^n} = a^{m-n}$:

$$= y^{1-4}$$ $$= y^{-3}$$

This can also be written as:

$$= \frac{1}{y^3}$$

The simplified expression is $\frac{1}{y^3}$.