For function $f(x)=\max_i x_i, x\in\mathbb{R}^n$. We define the Fenchel conjugate as $$ f^*(x^*) = \sup_x\Big\{ \langle x^*,x\rangle-f(x)\Big\}\,. $$
The standard subdifferential for $f(x)$ as $$ \partial f(x)=\Big\{x^*\in E: \forall y\in E\quad f(y)\geq f(x)+\langle x^*,y-x\rangle\Big\}\quad \text{if } x\in \operatorname{dom}(f)\,. $$ In the lecture note, it says
$$ x^*\in \partial f(x) \iff f^*(x^*)+f(x)=\langle x^*,x\rangle $$ I am confused with this statement.
I think it is simpler than you think and holds for general $f$.
When $$\tag{1} f^*(x^*)+f(x)=\langle x^*,x\rangle $$ then this is an $x$ in the domain of $f$ that attains the supremum in $$\tag{2} f^*(x^*) = \sup_{x}\Big\{\langle x^*,x\rangle-f(x)\Big\}\, $$ (without the braces this expression does not make sense).
Let $y\in E\,.$ Then it follows that $$\tag{3} \langle x^*,x\rangle-f(x)\ge\langle x^*,y\rangle-f(y)\,, $$ or $$\tag{4} f(y)\ge f(x)+\langle x^*,y-x\rangle\,. $$ That is, $x^*\in\partial f(x)\,.$ Conversely, assume $x^*\in\partial f(x)\,.$ Then (4) and (3) hold for all $y\in E\,.$ But this means $x$ is the one where $f$ attains the supremum in (2). Therefore (1) holds.
