The exercise (from Sakai) is:
Let $f: E\subseteq \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ be smooth and let $M_f := \{p = (x, f(x)) \in \mathbb{R}^n\,;\,x \in E\}$ be the graph of $f$ considered as a hypersurface of $\mathbb{R}^n$. Then, if $Df(x) = 0$ at $x \in E, \partial_n$ is a unit normal vector to $M_f$ at $x$. The second fundamental form of $M_f$ at $x$ is given by $\langle S (\partial_i , \partial_j), \partial_n \rangle = \partial_i \partial_j f$. Using this fact, let $p$ be a point in a Riemannian manifold $M$, $u \in U_p M$, and $A$ a symmetric linear transformation of $u^{\perp} := \{v \in T_p M \,;\,\langle u, v \rangle = 0\}$. Show that there exists a hypersurface $N$ of $M$ around $p$ with unit normal $u$ at $p$ such that the shape operator $A_u$ is equal to the given $A$.
My thought is to let $\text{dim}M=m$ and let $\tilde{M}$ be a submanifold of dimension $m-1$ containing $p$. Then there exists (I hope) $f \in C^{\infty}(\tilde{M})$ with $Df(p)=0$, so $N$ would be the graph of $f$ considered as a hypersurface. If that works, how do I go about showing that $A=A_u$? Is it sufficient to show that $\langle A_u v, w\rangle = \langle A(v,w),u \rangle = v_i v_j \frac{\partial^2 f}{\partial \xi^i \partial \xi^j}(p)$ for some $v,w \in u^{\perp}$ and coordinates $(\xi^i)$ of $M$?
Many thanks in advance!