Is that your work there? Because that looks good to me (so long as we agree that x is positive and that y and z are nonzero).

You are correct when you say that raising a product to an nth power is the same as multiplying that product by itself n times. The result of this is that the power gets "distributed" to each factor. That is, in general,

$\displaystyle (ab)^n=\overbrace{(ab)(ab)(ab)\cdots(ab)}^{n\ \mathrm{factors}}=a^nb^n$.

Another useful property is

$\displaystyle (a^m)^n = a^{m\cdot n} = a^{mn}$.

So,

$\displaystyle \frac{27z^9\left(y^2x^{-1/2}\right)^2}{\left(3z^3x^{-1/3}\right)^3y^4}$

$\displaystyle =\frac{27z^9\left(y^2\right)^2\left(x^{-1/2}\right)^2} {\left(3\right)^3\left(z^3\right)^3\left(x^{-1/3}\right)^3y^4}$

$\displaystyle =\frac{27z^9y^4x^{-1}}{27z^9x^{-1}y^4} = 1$,

again with $\displaystyle x>0$ and $\displaystyle y,z\neq0$.

Edit: But be sure you see that we cannot distribute powers over addition or subtraction. That is, $\displaystyle (a+b)^n\neq a^n+b^n$ for arbitrary $\displaystyle a$ and $\displaystyle b$ (though there are specific values for which this will be true). In this case, we would have to multiply $\displaystyle (a+b)$ by itself $\displaystyle n$ times.