Suppose we have a function $f : \mathbb{R}^2 \to \mathbb{R}$ whose graph is $(x, y, f(x,y))$. Let the parametric curve $\gamma(t) = (x(t),y(t),f(x(t),y(t)))$ be a level set (level curve) of $f$, that is $f(x(t),y(t)) = c$. Can we find a metric tensor field $g_{ij}$ defined on the graph of $f$ such that the curves $\gamma(t)$ are its geodesics?
From what I read in surface geometry (from Shifrin, Differential Geometry: A First Course in Curves and Surfaces), I would say no, here is my reasoning: at a point $p$ on a geodesic, the curvature vector $\kappa\mathbf{N}$ only has a component along $\mathbf{n}$, the outward normal to the graph at $p$, where $\mathbf{N} = \gamma'(t)/\Vert \gamma'(t) \Vert$ is the normal vector to the curve. Since a level curve is the intersection of the graph with a plane $z = c$, its normal vector $\mathbf{N}$ is horizontal, and thus $\mathbf{n}$ would have to be horizontal for the curve to be geodesic. Since the surface is the graph of a function, this cannot happen, and thus a level curve on a graph cannot be a geodesic.
Do this explanation make sense? I don't have a strong background in differential geometry, so please let me know if I'm way off. Thanks!