Simplify the first term inside the bracket, ( (16)/(9) )^-(3)/(2).

Using the exponent rule $a^{-n} = \frac{1}{a^n}$ and $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$:

$$\left( \frac{16}{9} \right)^{-\frac{3}{2}} = \left( \frac{9}{16} \right)^{\frac{3}{2}}$$

Now, using the rule $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$:

$$\left( \frac{9}{16} \right)^{\frac{3}{2}} = \left( \sqrt{\frac{9}{16}} \right)^3 = \left( \frac{\sqrt{9}}{\sqrt{16}} \right)^3 = \left( \frac{3}{4} \right)^3$$ $$\left( \frac{3}{4} \right)^3 = \frac{3^3}{4^3} = \frac{27}{64}$$

Using the exponent rule $a^{-n} = \frac{1}{a^n}$:

$$16^{-\frac{3}{4}} = \frac{1}{16^{\frac{3}{4}}}$$

Now, using the rule $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$:

$$\frac{1}{16^{\frac{3}{4}}} = \frac{1}{(\sqrt[4]{16})^3}$$

Since $2^4 = 16$, $\sqrt[4]{16} = 2$:

$$\frac{1}{(2)^3} = \frac{1}{8}$$

$$\left( \frac{27}{512} \right)^{\frac{1}{3}}$$

Using the rule $a^{\frac{1}{n}} = \sqrt[n]{a}$:

$$\left( \frac{27}{512} \right)^{\frac{1}{3}} = \sqrt[3]{\frac{27}{512}} = \frac{\sqrt[3]{27}}{\sqrt[3]{512}}$$

Since $3^3 = 27$ and $8^3 = 512$:

$$\frac{\sqrt[3]{27}}{\sqrt[3]{512}} = \frac{3}{8}$$

The simplified expression is $\frac{3}{8}$.
Comparing this with the given options:
A. $\frac{3}{4}$
B. $\frac{9}{16}$
C. $\frac{3}{8}$
D. $\frac{1}{4}$

The correct option is C.

The final answer is $\boxed{\text{C. } \frac{3}{8}}$.