Using the symmetry of the standard normal distribution about $0$, if $X \sim N(0, 1)$, then $-X \sim N(0, 1)$ as well, so $$ \begin{aligned} E\big[|X|X\big] &= P(X > 0) E\big[|X|X \,\big|\, X > 0\big] + P(X \leq 0) E\big[|X|X \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X^2 \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[\left(-X\right)^2 \,\big|\, -X \geq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X \,\big|\, X \geq 0\big] \\ &= 0. \end{aligned} $$

This generalizes to the fact that if $f : \mathbb{R} \to \mathbb{R}$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x \in \mathbb{R}$) and $X$ is a random variable with a distribution that is symmetric about zero, then $E[f(X)] = 0$, provided the expectation exists.

Using the symmetry of the standard normal distribution about $0$, if $X \sim N(0, 1)$, then $-X \sim N(0, 1)$ as well, so $$ \begin{aligned} E\big[|X|X\big] &= P(X > 0) E\big[|X|X \,\big|\, X > 0\big] + P(X \leq 0) E\big[|X|X \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X^2 \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[\left(-X\right)^2 \,\big|\, -X \geq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X^2 \,\big|\, X \geq 0\big] \\ &= 0. \end{aligned} $$

This generalizes to the fact that if $f : \mathbb{R} \to \mathbb{R}$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x \in \mathbb{R}$) and $X$ is a random variable with a distribution that is symmetric about zero, then $E[f(X)] = 0$, provided the expectation exists.