# Where does the $\ell_2$ norm subdifferential come from?

Looking at the definition of the subdifferential, we have that $$v$$ is a subdifferential of a function $$f$$ at a point $$x$$ if

$$f(y) \geq f(x) + g^T(y-x), \forall y$$

Now, for $$f(x) = \|x\|_2$$, it's not clear to me how the result is derived for when $$x\neq 0$$, which is that $$\partial f(x) = \frac{x}{\|x\|_2}$$.

Before we answer your question, let me be a bit more rigorous with the definitions: Let $$f:\mathbb{R}^n \to \mathbb{R}$$ be a proper convex function. Then, $$g\in \mathbb{R}^n$$ is said to be a subgradient of $$f$$ at $$x$$ if $$f(y) \ge f(x) + g^T(y-x)$$ for all $$y\in \mathbb{R}^n$$. The subdifferential of $$f$$ at $$x$$ is, instead, the set of all its subgradients at that point: $$\partial f(x) = \bigl\{g\in \mathbb{R}^n : g \text{ is a subgradient of f at x}\bigr\}.$$ Note the difference between subgradient and subdifferential. Also, note that these definitions hold only for functions that are convex (and proper, but this is only a technicality to avoid some degenerate cases).
Now, if $$f$$ is differentiable at $$x$$, then the gradient $$\nabla f(x)$$ is the only subgradient of $$f$$ at $$x$$ and the subdifferential is a singleton: $$\partial f(x) = \bigl\{\nabla f(x)\bigr\}.$$ The fact that the gradient is a subgradient is a result of $$f$$ being convex. To show that the gradient, if it exists, is the unique subgradient, you have to look at what happens in the limit for $$y\to x$$.
Wrapping up, for the $$\ell_2$$ norm we have (see also https://math.stackexchange.com/a/4335217/1218593) $$\partial f(x) = \begin{cases} \bigl\{g \in \mathbb{R}^n : \lVert g\rVert_2 \le 1\bigr\} &\text{if } x=0 \\ \bigl\{\nabla f(x)=\frac{1}{\lVert x\rVert_2}x\bigr\} &\text{if } x\ne 0. \end{cases}$$