If the function $$f(x) = \begin{cases} \dfrac{\log_e(1-x+x^2) + \log_e(1+x+x^2)}{\sec x - \cos x}, & x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) - \{0\} \\ k, & x = 0 \end{cases}$$ is continuous at $$x = 0$$, then $$k$$ is equal to:

We need to find $$k$$ such that the given piecewise function is continuous at $$x = 0$$, i.e., $$k = \displaystyle\lim_{x \to 0} \frac{\log_e(1-x+x^2) + \log_e(1+x+x^2)}{\sec x - \cos x}$$.

Using the logarithm property $$\log a + \log b = \log(ab)$$:

$$\log_e(1-x+x^2) + \log_e(1+x+x^2) = \log_e\left((1-x+x^2)(1+x+x^2)\right)$$

Expanding the product:

$$(1-x+x^2)(1+x+x^2) = (1+x^2)^2 - x^2 = 1 + 2x^2 + x^4 - x^2 = 1 + x^2 + x^4$$

So the numerator = $$\log_e(1 + x^2 + x^4)$$.

$$\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$$

$$\lim_{x \to 0} \frac{\log_e(1 + x^2 + x^4)}{\dfrac{\sin^2 x}{\cos x}} = \lim_{x \to 0} \frac{\log_e(1 + x^2 + x^4) \cdot \cos x}{\sin^2 x}$$

As $$x \to 0$$, using the standard limit $$\log_e(1+u) \approx u$$ for small $$u$$:

$$\log_e(1 + x^2 + x^4) \approx x^2 + x^4 \quad (\text{since } x^2 + x^4 \to 0)$$

Also, as $$x \to 0$$: $$\sin x \approx x$$, so $$\sin^2 x \approx x^2$$, and $$\cos x \to 1$$.

Substituting these approximations:

$$\lim_{x \to 0} \frac{(x^2 + x^4) \cdot 1}{x^2} = \lim_{x \to 0} (1 + x^2) = 1$$

We can verify using L'Hopital's rule. As $$x \to 0$$, both numerator and denominator approach 0 (it is a $$\frac{0}{0}$$ form since $$\log_e(1) = 0$$ and $$\sec 0 - \cos 0 = 0$$).

For continuity at $$x = 0$$, we need $$k = 1$$.

Therefore, the correct answer is Option A: 1.