I have the exercise to show for a Riemannian manifold $M$ and Levi-Civita-con $\nabla$, that
$g(\operatorname{grad}_gf,\operatorname{grad}_gf)=c\in \mathbb{R}$ implies $\nabla_{\operatorname{grad}_gf}\operatorname{grad}_gf =0$
Now I use that for $\operatorname{Hess}_g(f)(X,Y):= X(Y(f))-(\nabla_X Y)(f)$ it holds:
1) $\operatorname{Hess}_g(f)(X,Y)=\operatorname{Hess}_g(f)(Y,X)$ and
2) $\operatorname{Hess}_g(f)(X,Y)=g(\nabla_X\operatorname{grad}_gf,Y)$
What I have so far: I want to show that $g(\nabla_{\operatorname{grad}_gf}\operatorname{grad}_gf,X)=0$ for all $X$.
$g(\nabla_{\operatorname{grad}_gf}\operatorname{grad}_gf,X)=\operatorname{Hess}_g(f)(\operatorname{grad}_gf,X) (b)= \operatorname{Hess}_g(f)(X,\operatorname{grad}_g f) (a)= X(\operatorname{grad}_gf(f))-\nabla_X \operatorname{grad}_g f(f)$
Now I use $\operatorname{grad}_gf(f)= g(\operatorname{grad}_gf,\operatorname{grad}_gf)=c$ which implies that
$X(\operatorname{grad}_gf(f))-\nabla_X \operatorname{grad}_g f(f)=-\nabla_X \operatorname{grad}_g f(f)$
But now I don't know how to show that $-\nabla_X \operatorname{grad}_g f(f)=0$.
Is my approach correct? And if yes, how can I show that $-\nabla_X \operatorname{grad}_g f(f)=0$?
$Yf=df\ Y=({\rm grad}\ f,Y)$ so that $Yf=0$ iff $Y\perp {\rm grad}\ f$
Here $(\nabla_X {\rm grad} f,{\rm grad}\ f)=0$ so that $ \nabla_X {\rm grad}\ f$ is orthogonal to ${\rm grad}\ f$.