$(M,g,\nabla)$ Riemannian, $f:M\to\mathbb{R}$ then $g(\nabla_{\nabla f}\nabla f,\nabla f) = 0 \Rightarrow \nabla_{\nabla f}{\nabla f} = 0$ Let $(M,g,\nabla)$ a Riemannian manifold, $f:M\to\mathbb{R}$ then if $g(\nabla f,\nabla f) = 1$, then $\nabla_{\nabla f}{\nabla f} = 0.$
I can't follow from here:
$$\nabla_{\nabla f}g(\nabla f, \nabla f) = 2 g(\nabla_{\nabla f}{\nabla f},\nabla f) = 0.$$
How can I conclude that $\nabla_{\nabla f}{\nabla f} = 0?$
 A: Note that ${\rm Hess}\ f (v,w)=g(\nabla_v {\rm grad}\ f,w)$ In further ${\rm Hess}\ f$ is symmetric so that we assume that ON $\{E_i\}$ diagonalizes it 
By computation in OP, we can assume that ${\rm grad}\ f =E_n$ Hence ${\rm Hess} f (E_n,E_i)=0$ for all $i$ so that we complete the proof 
A: Observe that the value of $\nabla_XY(p)$ depends on the value of $X(p)$ and the value of $Y$ along a curve, tangent to $X$ at $p$. 
Also observe that given a Riemannian manifold of dimension $n$ and let $p\in M$ there exists a nbd $U\subset M$ of $p$ and $n$ vector fileld $\{E_1,\cdots ,E_n\}$ s.t at $p$, $\nabla_{E_i}E_j(p)=0$. Basically this is called a geodesic frame.
Now with respect to this geodesic frame on $U$ let $\nabla f(u)= \sum_j c_j(u) E_j(u)$ where $c_k=E_k(f)$. Now observe that at $p$, $E_iE_j(f)(p)=E_jE_i(f)(p)$ ( because co-variant derivative with respect to each other at $p$ is $0$). $<\nabla f,\nabla f>=1$ implies $\sum_jc_j(u)^2=1$. This implies $E_i(\sum_jc_j^2)=2\sum_jc_jE_i(c_j) =0$.Now $\nabla_{\nabla f}\nabla f(p)=\sum_{i,j}c_i(p)E_i(c_j(p))E_j =\sum_{i,j}c_i(p)E_j(c_i)E_j(p)=0$ (last equation i.e interchanging $E_i$ and $c_j$ at $p$ comes from the observation).
