Consider a coordinate transformation $\mathbf{y}(\mathbf{x})$. To keep things simple let's work in 2 dimensions so that we have a transformation $(x_1,x_2)$ to $(y_1, y_2) = \big(y_1(x_1,x_2), y_2(x_1,x_2)\big)$. Suppose the inverse transformation exists where $(x_1,x_2) = \big(x_1(y_1,y_2),x_2(y_1,y_2)\big)$ (but it might be very complicated).
I believe the following result is true: $$ \mathbb{J} = \left[ \begin{matrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{matrix} \right] \ = \ \left[ \begin{matrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{matrix} \right]^{-1} $$ Which allows for a simple way to compute the derivatives $\frac{\partial x_j}{\partial y_{k}}$.
My question is, is there an analogous way to compute the second derivatives $\frac{\partial^2 x_j}{\partial y_{k}\partial y_{\ell}}$?
EDIT: Could it involve a $4\times 4$ matrix of the form $$ \left[ \begin{matrix} \frac{\partial \mathbb{J}}{\partial y_1} & \frac{\partial \mathbb{J}}{\partial y_2} \\ \frac{\partial \mathbb{J}}{\partial y_1} & \frac{\partial \mathbb{J}}{\partial y_2} \end{matrix} \right] \ ? $$