Let $$f(x) = \frac{x^2 - x}{x^2 + 2x}$$, $$x \neq 0, -2$$. Then $$\frac{d}{dx}\left[f^{-1}(x)\right]$$ (wherever it is defined) is equal to:

We are given the function $$ f(x) = \frac{x^2 - x}{x^2 + 2x} $$ for $$ x \neq 0, -2 $$. First, we simplify this function by factoring both the numerator and the denominator.

The numerator is $$ x^2 - x = x(x - 1) $$. The denominator is $$ x^2 + 2x = x(x + 2) $$. Substituting these factorizations, we get:

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

Since $$ x \neq 0 $$, we can cancel the $$ x $$ in the numerator and denominator:

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

Now, we need to find the derivative of the inverse function, $$ \frac{d}{dx} [f^{-1}(x)] $$. To do this, we first find the inverse function $$ f^{-1}(x) $$. Set $$ y = f(x) = \frac{x - 1}{x + 2} $$. We solve for $$ x $$ in terms of $$ y $$:

$$ y = \frac{x - 1}{x + 2} $$

Multiply both sides by $$ x + 2 $$:

$$ y(x + 2) = x - 1 $$

Expand the left side:

$$ yx + 2y = x - 1 $$

Bring all terms involving $$ x $$ to one side and constants to the other:

$$ yx - x = -1 - 2y $$

Factor out $$ x $$ on the left:

$$ x(y - 1) = -1 - 2y $$

Solve for $$ x $$:

$$ x = \frac{-1 - 2y}{y - 1} $$

Simplify by factoring out $$-1$$ in both numerator and denominator:

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

Thus, the inverse function is $$ f^{-1}(x) = \frac{1 + 2x}{1 - x} $$.

Next, we differentiate $$ f^{-1}(x) $$ with respect to $$ x $$. Let $$ g(x) = f^{-1}(x) = \frac{1 + 2x}{1 - x} $$. We use the quotient rule for differentiation: if $$ g(x) = \frac{u}{v} $$, then $$ g'(x) = \frac{u'v - uv'}{v^2} $$, where $$ u = 1 + 2x $$ and $$ v = 1 - x $$.

Compute the derivatives: $$ u' = \frac{d}{dx}(1 + 2x) = 2 $$ and $$ v' = \frac{d}{dx}(1 - x) = -1 $$.

Apply the quotient rule:

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

Simplify the numerator:

$$ 2(1 - x) - (1 + 2x)(-1) = 2 - 2x + (1 + 2x) $$

Because subtracting a negative is addition:

$$ = 2 - 2x + 1 + 2x = 3 $$

The $$ -2x $$ and $$ +2x $$ cancel out, leaving:

$$ g'(x) = \frac{3}{(1 - x)^2} $$

Therefore, $$ \frac{d}{dx} [f^{-1}(x)] = \frac{3}{(1 - x)^2} $$.

Comparing with the options:

A. $$ \frac{-1}{(1-x)^2} $$

B. $$ \frac{3}{(1-x)^2} $$

C. $$ \frac{1}{(1-x)^2} $$

D. $$ \frac{-3}{(1-x)^2} $$

Our result matches option B.

Hence, the correct answer is Option B.