I'm in the process of deriving a discrete adjoint equation.
I'm trying to find the derivative of the vector $\textbf{X}_{1}$ with respect to the vector $\textbf{X}_0$ but I am not able to,
$\textbf{X}_{1}=\big (\mathbb{E}(\textbf{X}_0) \big) ^{-1}\,\mathbb{F}\textbf{X}_0$
with $\mathbb{F}$ a constant square matrix and $\mathbb{E}$ a invertible square matrix such that,
$\mathbb{E}(\textbf{X}_0)=\mathbb{I}_q^T \bigg ( diag(\mathbb{I}_q \,\textbf{X}_0) \,\mathbb{D}_x^q\bigg ).$
$\mathbb{I}_q$ and $\mathbb{D}_x^q$ are rectangular constant matrices. $diag(\textbf{Y})$ is the diagonal matrix having as diagonal entries the elements of $\textbf{Y}$.
I know that, for any vector $\textbf{Y}$,
$\frac{\partial \mathbb{E}(\textbf{Y})\textbf{X}_0}{\partial \textbf{Y}}=\mathbb{I}_q^T \bigg ( diag(\mathbb{D}_x^q\,\textbf{X}_0) \,\mathbb{I}_q\bigg )$
and,
$\frac{\partial \big (\mathbb{E}(\textbf{Y}) \big) ^{-1}\,\mathbb{F}\textbf{X}_0}{\partial \textbf{X}_0}=\big (\mathbb{E}(\textbf{Y}) \big) ^{-1}\,\mathbb{F}.$
Where should I go from here to get $\frac{\partial \textbf{X}_{1}}{\partial \textbf{X}_{0}}$
First, let's replace your elaborate variables with something easier to type $$\eqalign{ x &= \mathbf{X}_0 &,\,\,y &= \mathbf{X}_1 \cr E &= \mathbb{E}(\mathbf{X}_0) &,\,\,F &= \mathbb{F} \cr J &= \mathbb{I}_q &,\,\,K &= \mathbb{D}^q_x \cr }$$ Now your basic equation can be expressed as $$Ey = Fx$$ Taking derivatives (with respect to $x$) yields $$\eqalign{ dE\,y + E\,dy &= F\,dx \cr J^T{\rm diag}(Ky)J+E\,dy &= F\,I \cr E\,dy &= F - J^T{\rm diag}(Ky)J \cr dy &= E^{-1}\Big(F - J^T{\rm diag}(Ky)J\Big) \cr dy &= E^{-1}\Big(F - J^T{\rm diag}(KE^{-1}Fx)J\Big) \cr }$$ Reverting to your variables, yields the result in my previous comment.