For this question, if I divide this function into two parts, which are $x \ge 0$ and $x<0$, then the part that doesn't include "0" will have no critical point, and I also have no idea of how to find the singular point (might be the sharp point). Can anyone help me, thanks.
Recall that a crititcal point $p$ satisfies $\nabla f(p) = 0$ and that $\nabla f$ blows up at a singular point.
Suppose x < 0 so that $|x| = -x$. We then have $$\nabla f(x,y,z) = \Bigg(-3x^2, \quad y\Big(2-\frac{1}{\sqrt{z^2+y^2}}\Big) + 1, \quad \frac{-z}{\sqrt{z^2+y^2}}\Bigg) = 0,$$ from which we see that $x = z = 0$. This forces $y(2-\frac{1}{|y|}) + 1 = 0$.
For $y < 0$ we get $y = -1$ and $y = \frac{1}{2}$. But since we assume that $y<0$, $y = \frac{1}{2}$ does not apply.
For $y > 0$ we get $2y^2 + y + 1 = 0$. Noting that the left side has a negative determinant we conclude there are no solutions in $\mathbb{R}$ in this case.
Note that our presumption $x < 0$ in order to evaluate the absolute value $|x|$ was unnecessary since $x = 0$ from our condition $\nabla f(x,y,z) = 0$.
We have the critical point $A = (0,-1,0)$.
$\nabla f$ blows up at $(x,0,0)$ for all $x$, so we have the the set of singular points $B = (x,0,0)$.