Clarification: in either example, at the corner or the cusp, the tangent line is not defined. You have to go back to the definition of the derivative, which is the name of the slope of the tangent line.
You begin by writing down the slope of the secant line connecting the point $(0, 0)$ and a nearby point $\bigl( h, f(h) \bigr)$ on the graph of the function $f$. Imagine that $h$ is very small, but could b either positive or negative, as we’re about to ask what happens to the slope in the limit as $h$ goes to $0$. The slope, which depends on $h$ is
$$
m = \frac{\Delta y}{\Delta x} = \frac{f(h) - f(0)}{h - 0} = \frac{f(h)}{h}.
$$
Now, to obtain the derivative, we ask what this quantity approaches as $h \to 0$. (Notice that you can’t just plug in $h=0$.) In each of your two examples, this limit does not exist, so the derivative does not exist, so we say that the function is not differentiable at $x=0$. But the issue is slightly different in each case.
Corner: $f(x) = \lvert x \rvert$. Here, you find that for $h>0$, the slope is $m=1$ (independent of the magnitude $h$), but for $h<0$, we have $m=-1$. So, no matter how small we make $h$ in magnitude, these two values remain apart. In notation,
$$
\lim_{h \to 0^+} \frac{f(h)}{h} = +1
\neq -1 = \lim_{h \to 0^-} \frac{f(h)}{h}
\quad \Longrightarrow \quad
f’(0) = \lim_{h \to 0^+} \frac{f(h)}{h}
\text{ does not exist}.
$$
Cusp: $f(x) = \lvert x \rvert^{2/3}$. Here you find that for $h>0$, the slope is $m = h^{-1/3}$, a quantity that grows (positively) without bound as $h$ vanishes. On the other hand, for $h<0$, we have $m = -h^{-1/3}$, which grows (negatively) without bound. In limit notation,
$$
\lim_{h \to 0^+} \frac{f(h)}{h} = +\infty
\neq -\infty = \lim_{h \to 0^-} \frac{f(h)}{h}
\quad \Longrightarrow \quad
f’(0) = \lim_{h \to 0^+} \frac{f(h)}{h}
\text{ does not exist}.
$$
By the way, it might be useful to go through this process with a function such as $f(x) = x^2 + x$ at $x=0$, where the limit does exist, so the slope of the tangent line is well-defined $(m=1)$.
