How to show that $\left\{x^*\in X \mid \limsup_{y \to x,\;\|y\|=1}\frac{\langle x^*,y-x\rangle}{\|y-x\|} \le 0\right\} = \mathbb R x$ if $\|x\|=1$

Question

Let $X$ be a Banach space with topological dual $X^\star$. Given an extended value function $f:X \to \mathbb R \cup \{+\infty\}$, defined its Fréchet subdifferential at a point $x \in X$ as

$$ \partial f(x): = \left\{x^\star \in X^* \mid \liminf_{y \to x}\frac{f(y)-f(x) - \langle x^\star,y-x\rangle}{\|y-x\|} \ge 0\right\}. $$

Now consider the special case where $X$ is a Hilbert space and define the function $f:X \to \mathbb R \cup \{+\infty\}$ by $f(x) = 0$ if $\|x\|=1$; $f(x) = +\infty$ else.

I read a paper and the authors claim that $\partial f(x) = \mathbb R x$ for all $x \in X$ with $\|x\|=1$.

Question. How was that computation arrived at ?

My attempt

The problem can be reduced to showing that

$$ \left\{x^\star \in X \mid \limsup_{y \to x,\;\|y\|=1}\frac{\langle x^\star,y-x\rangle}{\|y-x\|} \le 0\right\} = \mathbb R x, $$ but even the above seems rather complex to establish.