Let $\Gamma$ be a $C^2$ compact submanifold of $\mathbb{R}^n$. Consider the distance function $\delta$ from $\Gamma$. It is well known that, for sufficiently small $\varepsilon>0$, $\delta$ is $C^2$ on $\{ 0<\delta < \varepsilon\}$, and that it satisfies the eikonal equation
$$ \| \nabla \delta \| = 1, \qquad \text{with} \qquad \delta|_{\Gamma} = 0. $$
Now recall Bochner's formula, valid for a smooth function $u \in C^\infty(M)$ on a general Riemannian manifold. It reads:
$$ \frac{1}{2} \Delta\left( \| \nabla u\|^2 \right) = \nabla u \cdot \nabla \Delta u + \| \mathrm{Hess} (u) \|^2_{\mathrm{HS}} + \mathrm{Ric}(\nabla u, \nabla u). \qquad (\star)$$
Here $\| \cdot \|_{\mathrm{HS}}$ is the Hilbert-Schmidt norm and $\Delta$ is the Laplace-Beltrami operator. When we specify $(\star)$ to $M = \mathbb{R}^n$ and $u$ is a smooth solution of the eikonal equation, we have the following:
$$ \nabla u \cdot \nabla \Delta u = - \| \mathrm{Hess} (u) \|^2_{\mathrm{HS}}. \qquad (\star\star)$$
Observe that the r.h.s. requires only $C^2$ regularity of $u$, while the l.h.s., a priori, requires a further derivative ($\nabla u \cdot \nabla \Delta u$ is indeed the directional derivative of $\Delta u$ in the direction $\nabla u$).
Q: Is it true, even if $\delta$ is a priori only a $C^2$ solution of the eikonal equation, that $\Delta \delta$ admits a directional derivative in the direction of $\nabla \delta$? Or, in other words, is it true that $(\star\star)$ holds for $\delta$?
The same question indeed can be posed in the Riemannian setting.
P.S: A direct computation with the distance from the $C^2$ (but not $C^3$) surface in $\mathbb{R}^2$ given by $y= x^{5/2}$ seems to support this claim.
This could be a standard fact about the regularity of solutions of the eikonal equation, but I have not been able to find any reference on this precise point.