If the value $\lambda$ is a function of a tensor $\textbf{C}=[C_{ij}]$,
$$\lambda=\tilde{\lambda}(\textbf{C})$$
$$\textbf{C}=\tilde{\textbf{C}}(\lambda)$$
then let us define $\textbf{D}=[D_{ij}]$ such
$$D_{ij}=\dfrac{\partial\tilde{\lambda}}{C_{ij}}$$
and also $\textbf{E}$ such
$$\textbf{E}=\dfrac{\partial \tilde{\textbf{C}}}{\partial \lambda}$$
then how valid is it to say
$$\textbf{D}\textbf{E}=\textbf{I}$$
where $\textbf{I}$ is the identity matrix. If it is proved I appreciate any support.
All values are in $\mathbb R$.
Assume that somehow, we have calculated the gradients $\frac{\partial\lambda}{\partial C}$ and $\frac{\partial C}{\partial\lambda}$
Then consider the differentials of each variable $$\eqalign{ d\lambda &= \frac{\partial\lambda}{\partial C} : dC \cr dC &= \frac{\partial C}{\partial\lambda} \, d\lambda \cr }$$ where a colon represents the trace/Frobenius product, i.e. $\,\,A:B={\rm tr}(A^TB)$
Now substitute the first differential into the second, and vice versa $$\eqalign{ d\lambda &= \frac{\partial\lambda}{\partial C} :\Bigg(\frac{\partial C}{\partial\lambda} \, d\lambda\Bigg) = \Bigg(\frac{\partial\lambda}{\partial C} :\frac{\partial C}{\partial\lambda}\Bigg) \, d\lambda \cr dC &= \frac{\partial C}{\partial\lambda} \, \Bigg(\frac{\partial\lambda}{\partial C} : dC\Bigg) = \Bigg(\frac{\partial C}{\partial\lambda} \, \frac{\partial\lambda}{\partial C}\Bigg) : dC \cr }$$ The terms in parenthesis on the RHS must be the identity elements for the respective product operations. $$\eqalign{ \frac{\partial\lambda}{\partial C} :\frac{\partial C}{\partial\lambda} &= \frac{\partial\lambda}{\partial\lambda} = 1 \cr \frac{\partial C}{\partial\lambda} \, \frac{\partial\lambda}{\partial C} &= \frac{\partial C}{\partial C} = {\mathcal E} \cr }$$ where the 4th order tensor ${\mathcal E}$ has components $${\mathcal E}_{ijkl} = \delta_{ik}\,\delta_{jl}$$ There is no combination of these gradients which yields the identity matrix.
However, if we were considering a vector $v$ instead of a matrix $C$, then the identity matrix can be obtained as $$\eqalign{ \frac{\partial v}{\partial\lambda} \, \frac{\partial\lambda}{\partial v} &= \frac{\partial v}{\partial v} = I \cr }$$