What is the covariance between $X$ and $X^2$ (without assuming normality)?

If $$X \sim \text{N}(0, \sigma^2)$$ then it is well-known (from Cochran's theorem) that $$X \ \bot \ X^2$$ so these random variables have zero covariance. However, this holds only for the normal distribution with zero mean. What is the general formula for $$\mathbb{C}(X,X^2)$$ without assuming normality or zero mean?

The general form of the covariance depends on the first three moments of the distribution. To facilitate our analysis, we suppose that $$X$$ has mean $$\mu$$, variance $$\sigma^2$$ and skewness $$\gamma$$. The covariance of interest exists if $$\gamma < \infty$$ and does not exist otherwise. Using the relationship between the raw moments and the cumulants, you have the general expression:
\begin{aligned} \mathbb{C}(X,X^2) &= \mathbb{E}(X^3) - \mathbb{E}(X) \mathbb{E}(X^2) \\[6pt] &= ( \mu^3 + 3 \mu \sigma^2 + \gamma \sigma^3 ) - \mu ( \mu^2 + \sigma^2 ) \\[6pt] &= 2 \mu \sigma^2 + \gamma \sigma^3. \\[6pt] \end{aligned}
The special case for an unskewed distribution with zero mean (e.g., the centred normal distribution) occurs when $$\mu = 0$$ and $$\gamma = 0$$, which gives zero covariance. Note that the absence of covariance occurs for any unskewed centred distribution, though independence holds only for the normal distribution.