This is the exercise 12 on page 278 of Analysis II of Amann and Escher.
Suppose $X$ is open in $\Bbb R^n$ and $f\in C^1(X,\Bbb R)$. Also let $\nu:G\to\mathrm S^n$ be a unit normal on $G:={\rm graph}(f)$. Show that $\nu\in C^\infty(G,\mathrm S^n)$.
An unit normal $\nu(p)$, for some $p\in G$, is the tangent part of $(p,\nu(p))\in T_p^\bot G$, that it means that $(\nu(p)|w)=0$ whenever $(p,w)\in T_pG$.
Also $\nu$ is differentiable when there are charts $\varphi$ and $\psi$ of $G$ and $\mathrm S^n$ respectively for the same territories (that is, $\operatorname{im}(\nu\circ\varphi^{-1})\subset\operatorname{im}(\psi^{-1})$) such that $\nu_{\varphi,\psi}:=\psi\circ\nu\circ\varphi^{-1}$ is differentiable.
My work so far: if $\nu: G\to\mathrm S^n$ is an unit normal for $G:=\text{graph}(f)$, for some $f\in C^1(X,\Bbb R)$ with $X\subset\Bbb R^n$ open then we have that $$ \nu(x,f(x))=\frac{(-\nabla f(x),1)}{\sqrt{1+|\nabla f(x)|^2}} $$ because $(\nu(x_0,f(x_0))|w)=0$ for any $(p,w)\in T_pG$ where $p:=(x_0,f(x_0))$, because $T_pG:={\rm im}(T_{x_0} g)={\rm span}[(e_1,\partial_1 f(x_0)),\ldots,(e_n,\partial_n f(x_0))]$, for $g(x):=(x,f(x))$.
Now let $\varphi=\pi_1$, thus $g=\varphi^{-1}$, and $\varphi$ is a chart (and an atlas) of $G$. Also we will use the stereographic projections $\psi_\pm(y):=y'/(1\mp y_{n+1})$ where $y:=(y',y_{n+1})$ as an atlas of $\mathrm S^n$. Thus $$ \begin{align}\nu_{\varphi,\psi_\pm}(x)&=(\psi_\pm\circ\nu)(x,f(x))\\&=-\frac{\nabla f(x)}{\left(1\mp\frac1{\sqrt{1+|\nabla f(x)|^2}}\right)\sqrt{1+|\nabla f(x)|^2}}\\&=-\frac{\nabla f(x)}{\sqrt{1+|\nabla f(x)|^2}\mp 1}\end{align} $$ However it is not clear that this map is smooth if just $f\in C^1$. Maybe there is a typo in the exercise or there is a way to show that $\nu$ is smooth?