Let $$f(x) = \cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)$$, $$0 \lt x \lt 1$$. Then:

We are given the function $$f(x) = \cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)$$ for $$0 < x < 1$$.

First let us simplify the innermost term by substituting $$\theta = \cot^{-1}\sqrt{\frac{1-x}{x}}$$.

This gives $$\cot \theta = \sqrt{\frac{1-x}{x}}$$.

From the definition of cotangent in a right triangle, the base is $$\sqrt{1-x}$$ and the perpendicular is $$\sqrt{x}$$.

The hypotenuse becomes $$\sqrt{(\sqrt{1-x})^2 + (\sqrt{x})^2} = \sqrt{1-x+x} = 1$$.

Therefore, the sine of this angle is $$\sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} = \sqrt{x}$$.

Substituting this back into the expression simplifies the function to $$f(x) = \cos\left(2\tan^{-1}\sqrt{x}\right)$$.

Next, let $$\phi = \tan^{-1}\sqrt{x}$$, which means $$\tan \phi = \sqrt{x}$$.

Using the double angle identity for cosine in terms of tangent, we have $$\cos(2\phi) = \frac{1 - \tan^2\phi}{1 + \tan^2\phi}$$.

Substituting $$\tan \phi = \sqrt{x}$$ gives $$f(x) = \frac{1 - x}{1 + x}$$.

Now we differentiate $$f(x)$$ with respect to $$x$$ using the quotient rule:

$$f'(x) = \frac{(-1)(1 + x) - (1 - x)(1)}{(1 + x)^2} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}$$

Now let us test the expression in option A, which is $$(1 - x)^2 f'(x) + 2(f(x))^2$$:

Substituting our values for $$f(x)$$ and $$f'(x)$$, we get:

$$(1 - x)^2 \left(\frac{-2}{(1 + x)^2}\right) + 2\left(\frac{1 - x}{1 + x}\right)^2$$

$$= \frac{-2(1 - x)^2}{(1 + x)^2} + \frac{2(1 - x)^2}{(1 + x)^2} = 0$$

This matches option A exactly.

Hence the correct option is A.