If $$S_1$$ and $$S_2$$ are respectively the sets of local minimum and local maximum points of the function, $$f(x) = 9x^{4} + 12x^{3} - 36x^{2} + 25$$, $$x \in R$$, then:

We have the quartic polynomial function $$f(x)=9x^{4}+12x^{3}-36x^{2}+25,\; x\in\mathbb{R}.$$

To locate its local extrema, we first need the first derivative. Using the rule $$\dfrac{d}{dx}\bigl(x^{n}\bigr)=nx^{\,n-1},$$ we differentiate term-by-term:

$$\begin{aligned} f'(x) &= \dfrac{d}{dx}\bigl(9x^{4}\bigr)+\dfrac{d}{dx}\bigl(12x^{3}\bigr)-\dfrac{d}{dx}\bigl(36x^{2}\bigr)+\dfrac{d}{dx}(25) \\ &= 9\cdot4x^{3}+12\cdot3x^{2}-36\cdot2x+0 \\ &= 36x^{3}+36x^{2}-72x. \end{aligned}$$

Now we factor $$f'(x)$$ completely. First we take out the common factor $$36x$$:

$$f'(x)=36x\bigl(x^{2}+x-2\bigr).$$

The quadratic $$x^{2}+x-2$$ factors further because its discriminant is $$1^{2}-4(1)(-2)=9=3^{2}.$$ So,

$$x^{2}+x-2=(x+2)(x-1).$$

Substituting this back,

$$f'(x)=36x(x+2)(x-1).$$

Critical points occur where $$f'(x)=0$$, i.e.

$$36x(x+2)(x-1)=0 \;\Longrightarrow\; x=0,\;x=-2,\;x=1.$$

Next, to determine the nature of each critical point, we use the Second Derivative Test. First, we compute the second derivative. Differentiating $$f'(x)=36x^{3}+36x^{2}-72x$$ once more, we apply the same power rule:

$$\begin{aligned} f''(x) &= \dfrac{d}{dx}\bigl(36x^{3}\bigr)+\dfrac{d}{dx}\bigl(36x^{2}\bigr)-\dfrac{d}{dx}\bigl(72x\bigr) \\ &= 36\cdot3x^{2}+36\cdot2x-72 \\ &= 108x^{2}+72x-72. \end{aligned}$$

The Second Derivative Test says: if $$f''(c)>0$$ the point $$x=c$$ is a local minimum, while if $$f''(c)<0$$ it is a local maximum.

We evaluate $$f''(x)$$ at each critical point:

1. For $$x=-2$$,

$$f''(-2)=108(-2)^{2}+72(-2)-72 =108\cdot4-144-72 =432-216 =216>0.$$

So $$x=-2$$ gives a local minimum.

2. For $$x=0$$,

$$ f''(0)=108\cdot0^{2}+72\cdot0-72 =-72<0. $$

Thus $$x=0$$ yields a local maximum.

3. For $$x=1$$,

$$ f''(1)=108(1)^{2}+72(1)-72 =108+72-72 =108>0. $$

Hence $$x=1$$ is also a local minimum.

Collecting our results, the set of local minima is $$S_{1}=\{-2,\,1\}$$ and the set of local maxima is $$S_{2}=\{0\}.$$

Comparing with the given options, this corresponds to Option D.

Hence, the correct answer is Option D.