It is easy to see that the domain of $f(x)$ is $(-\infty, 0) \cup (0, +\infty)$. When $x \in (-\infty, 0) \cup (0, +\infty)$, we have
$$\begin{aligned} f(-x) &= \frac{-x}{1-2^{-x}} - \frac{x}{2} \\ &= \frac{-x \cdot 2^x}{2^x - 1} + \frac{x}{2} \\ &= \frac{x}{1-2^x} - x + \frac{x}{2} \\ &= \frac{x}{1-2^x} - \frac{x}{2} = f(x). \end{aligned}$$
Therefore, $f(x)$ is an even function, and obviously not an odd function. Answer: A.