If $$f(x)$$ is continuous and $$f\left(\frac{9}{2}\right) = \frac{2}{9}$$, then $$\lim_{x \to 0} f\left(\frac{1-\cos 3x}{x^2}\right)$$ equals to:

We are given that $$ f(x) $$ is continuous and $$ f\left(\frac{9}{2}\right) = \frac{2}{9} $$. We need to find the limit:

$$ \lim_{x \to 0} f\left( \frac{1 - \cos 3x}{x^2} \right) $$

Since $$ f $$ is continuous, we can move the limit inside the function. This means:

$$ \lim_{x \to 0} f\left( \frac{1 - \cos 3x}{x^2} \right) = f\left( \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} \right) $$

So, we first compute the limit inside:

$$ L = \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} $$

We know the trigonometric identity $$ 1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right) $$. Substituting $$ \theta = 3x $$:

$$ 1 - \cos 3x = 2 \sin^2 \left( \frac{3x}{2} \right) $$

Therefore:

$$ \frac{1 - \cos 3x}{x^2} = \frac{2 \sin^2 \left( \frac{3x}{2} \right)}{x^2} $$

We can rewrite this as:

$$ \frac{2 \sin^2 \left( \frac{3x}{2} \right)}{x^2} = 2 \cdot \left( \frac{\sin \left( \frac{3x}{2} \right)}{x} \right)^2 $$

Now, express $$ \frac{\sin \left( \frac{3x}{2} \right)}{x} $$ in terms of $$ \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} $$:

$$ \frac{\sin \left( \frac{3x}{2} \right)}{x} = \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \cdot \frac{3}{2} $$

So:

$$ \left( \frac{\sin \left( \frac{3x}{2} \right)}{x} \right)^2 = \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \cdot \frac{3}{2} \right)^2 = \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 \cdot \left( \frac{3}{2} \right)^2 = \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 \cdot \frac{9}{4} $$

Substituting back:

$$ \frac{1 - \cos 3x}{x^2} = 2 \cdot \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 \cdot \frac{9}{4} = \frac{18}{4} \cdot \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 = \frac{9}{2} \cdot \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 $$

As $$ x \to 0 $$, let $$ t = \frac{3x}{2} $$. Then $$ t \to 0 $$. We know that:

$$ \lim_{t \to 0} \frac{\sin t}{t} = 1 $$

So:

$$ \lim_{x \to 0} \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} = 1 $$

Therefore:

$$ \lim_{x \to 0} \left( \frac{\sin \left( \frac{3x}{2} \right)}{\frac{3x}{2}} \right)^2 = 1^2 = 1 $$

Thus:

$$ L = \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \frac{9}{2} \cdot 1 = \frac{9}{2} $$

Now, substituting back into the original expression:

$$ \lim_{x \to 0} f\left( \frac{1 - \cos 3x}{x^2} \right) = f\left( \frac{9}{2} \right) $$

Given that $$ f\left( \frac{9}{2} \right) = \frac{2}{9} $$, we have:

$$ f\left( \frac{9}{2} \right) = \frac{2}{9} $$

Hence, the limit is $$ \frac{2}{9} $$.

Comparing with the options:

A. $$ \frac{8}{9} $$

B. 0

C. $$ \frac{2}{9} $$

D. $$ \frac{9}{2} $$

So, the correct answer is Option C.