Let $(M^2,g)$ be a compact connected Riemannian surface without boundary. We denote by $\Delta = -div(\nabla)$ the Laplace-Beltrami operator. Suppose $u$ and $v$ are eigenfunctions of $\Delta$ associated to the same eigenvalue $\lambda > 0$. If $\nabla u(p) = h(p) \nabla v(p)$ for some continuous function $h : M \to \mathbb{R}$ and for all $p \in M$, does it follow that $u$ is a constant multiple of $v$?
My thoughts: the function $h$ is smooth away from the (discrete) critical set $C$ of $u$ and $v$. On $M \setminus C$ we can apply the divergence operator on both sides of the above equation to get
$$\Delta u = -g(\nabla h, \nabla v) + h \Delta v.$$
Since $u$ and $v$ are eigenfunctions,
$$\lambda u= -g (\nabla h, \nabla v) + h \lambda v,$$
or
$$\lambda(u - hv) = - g(\nabla h, \nabla v).$$
How do I continue? Do you have other ideas?