Consider the function $ f:[0,1] \to \mathbb{R}, \qquad$ $ f(x) = \begin{cases} x^2 & \text{if $x > 0$,} \\ 1 & \text{if $x=0$.} \end{cases} $
This function is convex. I want to find subdifferential at $x = 0$. Does it exist? Is it $\phi$ or something else?
It's the empty set. You can visually see this from the epigraph; there's no non-vertical line supporting the epigraph at the point $(0, 1)$.
Applying the definition, suppose that $y \in \partial f(0)$. Then, for all $x \in (0, 1]$, we have, $$yx = \langle y, x - 0 \rangle \le f(x) - f(0) = x^2 - 1.$$ As $x \to 0^+$, the left hand side tends to $0$, while the right hand side tends to $-1$. This contradicts the inequality above; eventually $yx - (x^2 - 1) \ge 0$ for $x$ sufficiently close to $0$.