Let $F:R^n \times R^m \to R^m$ be a function. Suppose we have $F(x,y)=0$ where $y \in R^m$ is implicitly defined as a function of $x \in R^n$: $y=h(x)$. We have $F(x,h(x))=0$. By the Chain Rule, the Implicit Function Theorem can be derived: \begin{align*} &D_xF + D_yF Dh = 0 \\ & \implies Dh = -(D_yF)^{-1} D_xF \end{align*}By taking the derivative again, we obtain \begin{align*} \big( D_{xx}F + D_{xy}F Dh \big) + \big( D_{yx}F + D_{yy}F Dh \big) Dh + D_yF D^2h= 0 \end{align*} I want to solve for $D^2h$. In this case, what is the meaning of the 'inverse' of $D_yF$?
If $F: R \to R$, the solution is simple and explained here: Implicit Function Theorem second derivative calculation help
However, in my problem at hand, $h$ is necessarily a vector-valued function that takes a vector-argument. Does any one know a convenient way to express $D^2h$ (e.g. in a matrix or tensor format)?