If anybody can help with this. Given a point $\boldsymbol x = (x,y)$, it can be represented as $\boldsymbol x=\boldsymbol p(s)+r \boldsymbol u(s) $. p is a curve parametrized with arc length s. $r$ is the distance between $\boldsymbol p$ and $\boldsymbol x$. $\theta$ is the angle between the tangent vector of $\boldsymbol p$ and the x-axis. $\boldsymbol u(s)$ is the unit normal vector of $\boldsymbol p$ at s. How can I prove that
$$\begin{vmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial r} \end{vmatrix} = 1 + r \dfrac{d\theta}{ds} $$ I think that relates somehow to the determinant of jacobian of frenet frame but I could not find any resource that has the proof.
Writing the curvature $\kappa = d\theta/ds$ and suppressing the vector nature of the problem just makes it difficult. You are right that the Frenet equations are relevant.
We have $\dfrac{\partial x}{\partial s} = T + r(-\kappa T)$ (where $T,N$ are the Frenet frame of the curve $p$) and $\dfrac{\partial x}{\partial r} = N$. So the determinant should be $1-r\kappa$, not the $1+r\kappa$ you have.