Let $$f(x) = x^2 + \frac{1}{x^2}$$ and $$g(x) = x - \frac{1}{x}$$, $$x \in R - \{-1, 0, 1\}$$. If $$h(x) = \frac{f(x)}{g(x)}$$, then the local minimum value of h(x) is:

We have the two auxiliary functions $$f(x)=x^{2}+\dfrac{1}{x^{2}}$$ and $$g(x)=x-\dfrac{1}{x}$$ with the domain $$x\in\mathbb{R}\;-\;\{-1,0,1\}$$. The function whose local extrema we need is

$$h(x)=\dfrac{f(x)}{g(x)}=\dfrac{x^{2}+\dfrac{1}{x^{2}}}{\,x-\dfrac{1}{x}\,}.$$

First we simplify the numerator. We recall the algebraic identity

$$(a-b)^{2}=a^{2}-2ab+b^{2}.$$

Taking $$a=x$$ and $$b=\dfrac{1}{x}$$ gives

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

From this we immediately obtain

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

Substituting the expressions $$f(x)=g(x)^{2}+2$$ and $$g(x)=x-\dfrac{1}{x}$$ into $$h(x)$$ we arrive at

$$h(x)=\dfrac{g(x)^{2}+2}{g(x)}.$$

To avoid repeatedly writing the longer symbol, we now put $$t=g(x)=x-\dfrac{1}{x}.$$ The domain restriction $$x\neq -1,0,1$$ implies $$t\neq 0$$ (because $$x=1$$ or $$x=-1$$ would make $$t=0$$). Using this new variable the function simplifies beautifully to

$$h(x)=t+\dfrac{2}{t}\equiv\varphi(t).$$

So the original problem has been reduced to finding the local minimum of

$$\varphi(t)=t+\dfrac{2}{t},\qquad t\in\mathbb{R}\;-\;\{0\}.$$

To locate critical points we differentiate. The derivative of $$\dfrac{2}{t}$$ is $$-\dfrac{2}{t^{2}}$$, hence

$$\dfrac{d\varphi}{dt}=1-\dfrac{2}{t^{2}}.$$

Critical points occur where the derivative vanishes:

$$1-\dfrac{2}{t^{2}}=0\;\Longrightarrow\;t^{2}=2\;\Longrightarrow\;t=\sqrt{2}\;\text{or}\;t=-\sqrt{2}.$$

To decide the nature of these points we look at the second derivative. Differentiating once more, we obtain

$$\dfrac{d^{2}\varphi}{dt^{2}}=\dfrac{4}{t^{3}}.$$

Evaluating at the two critical values gives

$$\dfrac{d^{2}\varphi}{dt^{2}}\Bigg|_{t=\sqrt{2}}=\dfrac{4}{(\sqrt{2})^{3}}=\dfrac{4}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}\;>\;0,$$

$$\dfrac{d^{2}\varphi}{dt^{2}}\Bigg|_{t=-\sqrt{2}}=\dfrac{4}{(-\sqrt{2})^{3}}=-\dfrac{2}{\sqrt{2}}\;<\;0.$$

Because the second derivative is positive at $$t=\sqrt{2}$$, the point $$t=\sqrt{2}$$ gives a local minimum, while $$t=-\sqrt{2}$$ gives a local maximum.

We now compute the minimum value itself:

$$\varphi(\sqrt{2})=\sqrt{2}+\dfrac{2}{\sqrt{2}}=\sqrt{2}+\dfrac{2\sqrt{2}}{2}= \sqrt{2}+\sqrt{2}=2\sqrt{2}.$$

Finally, we verify that this minimum actually corresponds to some admissible $$x$$. Setting $$x-\dfrac{1}{x}=\sqrt{2}$$ and solving, we get

$$x^{2}-\sqrt{2}\,x-1=0\;\Longrightarrow\;x=\dfrac{\sqrt{2}\pm\sqrt{6}}{2},$$

neither of which equals $$-1,0,1$$, so the critical value is indeed attained within the domain.

Therefore the local minimum value of $$h(x)$$ is $$2\sqrt{2}$$.

Hence, the correct answer is Option A.