Differentiability of $f(x)=|x^3|$ at $x=0$ via the limit definition

For $h > 0$, $f(h) = |h^3| = h^3$, so $\frac{f(h)-f(0)}{h} = \frac{h^3}{h} = h^2$, which tends to $0$ as $h \to 0^+$. For $h < 0$, $f(h) = |h^3| = -h^3$, so $\frac{f(h)-f(0)}{h} = \frac{-h^3}{h} = -h^2$, which also tends to $0$ as $h \to 0^-$.

Since both one-sided limits equal $0$, the two-sided limit exists and equals $0$. Therefore $f(x) = |x^3|$ is differentiable at $x=0$ with $f'(0) = 0$; unlike $|x|$, cubing smooths out the corner so no sharp point or cusp remains at the origin.

No, it does not have a sharp corner at zero. As shown in the earlier part, the left-hand and right-hand difference quotients for absolute value of x cubed both approach zero, so the derivative exists there. This means the graph has a smooth, flat tangent line at the origin rather than a corner. The cubing operation flattens out the sharpness that absolute value of x exhibits at that same point.