Denote $g_{ij}$ to be the metric and $(g^{-1})^{ij}$ to be its inverse. Let $u$ and $\underline{u}$ satisfy eikonal equations, i.e. $$(g^{-1})^{\mu\nu}\partial_{\mu}u\partial_{\nu}u=0,\quad(g^{-1})^{\mu\nu}\partial_{\mu}\underline{u}\partial_{\nu}\underline{u}=0.$$
Define the vector fields $L'$ and $\underline{L'}$ by
$$ L'^{\mu}=-2(g^{-1})^{\mu\nu}\partial_{\nu}u,\quad \underline{L}'^{\mu}=-2(g^{-1})^{\mu\nu}\partial_{\nu}\underline{u}.$$
Then,
$$\nabla_{L'}L'=0, \quad\nabla_{\underline{L}'}\underline{L}'=0.$$
I don't know how to show the last two are vanishing. Can anyone possibly help please. Every suggestion is really appreciated.
Here’s a general calculation we can make. Suppose $(M,g)$ is a semi-Riemannian manifold, and $\nabla$ is the Levi-Civita connection, and $f:M\to\Bbb{R}$ is any smooth function. This function gives rise to a covector field $df$, which via the metric gives rise to a vector field $\text{grad}(f)$. Now, for any smooth vector fields $X,Y$, we can construct the function \begin{align} g\left(\nabla_X(\text{grad}(f)),Y\right). \end{align} I claim that this expression is symmetric with respect to $X$ and $Y$ (i.e $\nabla(\text{grad}(f))$ is self-adjoint relative to $g$). The proof uses the metric isomorphism $\flat\equiv g^{\flat}$ (i.e index lowering) and the fact that it commutes with covariant derivatives (this is the metric-compatibility part of Levi-Civita connection), and that the Hessian of smooth functions are symmetric (this is due to the torsion-freeness of the Levi-Civita connection). So, here we go: \begin{align} g\left(\nabla_X(\text{grad}(f)),Y\right)&=[\nabla_X(\text{grad}(f))]^{\flat}(Y)\tag{definition of $\flat$}\\ &=\left(\nabla_X\left([\text{grad}(f)]^{\flat}\right)\right)(Y)\tag{metric-compatibility}\\ &=\left(\nabla_X(df)\right)(Y)\tag{definition of grad}\\ &=\left(\nabla_X(\nabla f)\right)(Y)\tag{definition of $\nabla$}\\ &=(\nabla\nabla f)(X,Y). \end{align} This final expression is symmetric in $X,Y$ since $f$ is a smooth function and $\nabla$ is torision-free. Therefore, \begin{align} g\left(\nabla_X(\text{grad}(f)),Y\right)= g\left(\nabla_Y(\text{grad}(f)),X\right).\tag{$*$} \end{align} Now, let us specialize to the case where $L:=\text{grad}(f)$ is such that $g(L,L)$ is constant (which is certainly true if $f=u$ solves the Eikonal equation). By choosing $X=L$ and keep $Y$ arbitrary, we get \begin{align} g\left(\nabla_LL,Y\right)&=g\left(\nabla_YL,L\right)=\frac{1}{2}\nabla_Y\left(g(L,L)\right)=0, \end{align} where the first equality used $(*)$, then we used the product rule (and metric compatibility of course), and finally that $g(L,L)$ is constant. Since this expression vanishes for all $Y$, the non-degeneracy of the metric implies that $\nabla_LL=0$. You can play the same game with $\underline{L}$.