I'm stuck in the second part (the inequality part) of the following: let $S$ be a compact regular surface, $f: S \longrightarrow \mathbb{R}$ given by $f(p)=\langle p,p \rangle$, and $p_0 \in S$ a maximum point of $f$. Show that $T_{p_0}S=\{p_0\}^\perp$, and that, relatively to the orientation $N$ of $S$ for which $N(p_0)=-p_0/\|p_0\|$, the following inequality holds $$II_{p_0}(w) \geq \frac{1}{\|p_0\|}, w \in T_{p_0}S, \|w\|=1 $$ In which $II_{p_0}(w)=\langle-dN_{p_0}(w),w \rangle$ is the Second Fundamental Form of $S$ at $p_0$.
Cauchy-Schwarz comes to mind, but it seems that I would only get the reverse inequality, which kind of does not make sense. Besides, to directly compute $dN_{p_0}$ I would need to know what values the orientation N takes around $p_0$, no?
Phrased slightly differently, this question appears many times on MSE. Choose a curve $\gamma(t)$ in $S$ with $\gamma(0)=p_0$ and $\gamma'(0)=w$. Then the function $F=f\circ\gamma$ has maximum at $t=0$. It follows that $F'(0)=0$ and $F''(0)\le 0$. Since $F'(t)=Df_{\gamma(t)}(\gamma'(t))=2\langle\gamma(t),\gamma'(t)\rangle$, we have $$0\ge\tfrac12 F''(0)=\langle\gamma'(0),\gamma'(0)\rangle + \langle p_0,\gamma''(0)\rangle = 1+\langle p_0,\gamma''(0)\rangle,$$ so $$\langle p_0,\gamma''(0)\rangle\le -1.$$ Note that $p_0 = -\|p_0\|N(p_0)$, as you said. Rewriting, we have $$\|p_0\|\langle N(p_0),\gamma''(0)\rangle \ge 1.$$
All we need now is the standard product rule manipulation that relates normal curvature of the curve to the second fundamental form. Differentiating $\langle N(\gamma(t)),\gamma'(t)\rangle = 0$ gives $$\langle N(p_0),\gamma''(0)\rangle = -\langle (N\circ\gamma)'(0),\gamma'(0)\rangle = -\langle dN_{p_0}(w),w\rangle = \text{II}_{p_0}(w).$$ Therefore, $\text{II}_{p_0}(w)\ge \dfrac1{\|p_0\|}$, as desired.
(COMMENT: The text did specify the orientation of the normal vector. So I'm not sure what your last sentence means.
P.S. You might find my free differential geometry text of value.)