 Solution

The one-dimensional kernel of $d{K}_x$ is the tangent to its level curve; the metric gives two unit normals and $d{K}_x(e_x)>0$ selects one. Isometries preserve the metric, $K$, and this orientation choice, so they carry $e_x$ to $e_{\varphi (x)}$. Therefore $g(x)=d{K}_x(e_x)=|\nabla K(x)|$ is isometry invariant.