Let $\{e_1,...,e_n\}$ be standard basis of $\mathbb R^n$, consider the cylinder $$L=\{x\in \mathbb R^n : |x-\langle x, e_1\rangle e_1|=1 \}$$ Obviously, mean curvature of $L$ is $H=n-1$. Assume $\{x_1,..., x_{n-1}\}$ is a local coordinate of $L$, then , the gradient of $H$ is $$\nabla H = g^{ij} \frac{\partial H}{\partial x_i}\frac{\partial}{\partial x_j}=0$$ On the other hand, the normal vector of $L$ is $$\nu=x-\langle x, e_1\rangle e_1$$ And $$\langle x, \nu\rangle =1$$ So, we have $$H=(n-1)\langle x, \nu \rangle$$ Then, $$\nabla \langle x, \nu \rangle =0$$ Now I want to verify $\nabla \langle x, \nu \rangle =0$.
$$\nabla \langle x, \nu \rangle =g^{ij}\frac{\partial \langle x, \nu \rangle }{\partial x_i}\frac{\partial}{\partial x_j} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g^{ij} (\langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle) \frac{\partial}{\partial x_j}$$ It is equal to show $$\langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle =0$$ Because $\nabla_i x$ is tangent vector, $\langle \nabla_i x, \nu\rangle=0$. Then $$\langle x, \nabla _i \nu \rangle = \langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle$$ If the local coordinate is $$x_1=e_1, ~~~~\{x_2,...,x_n\} \text{ is the spherical coordinates }$$ Then, we have g_{ij}=\delta_{ij} ~~~\text{ and }~~~ h_{ij}=\left\{ \begin{aligned} &1 ~~~~~i=j\ne 1 \\ &0 ~~~~~\text{others}\\ \end{aligned} \right. Then $$\langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle=0$$ Since $$\langle x, \frac{\partial x}{\partial x_i} \rangle =0 ~~~\text{when i\ne 1,} ~~~~~\text{and}~ h_{11}=0$$
PS: At beginning , I don't know how to calculate $\nabla_i x$ and $\nabla _i \nu$. Then, I know , and add my calculate here.