Derivative of function with the Kronecker product of a Matrix with respect to vech I have $\Sigma$ a symmetric $2 \times 2$ matrix, and $\Sigma^{-1}$ is its inverse. 
Now, $\tilde{\Sigma}^{-1}=\Sigma^{-1} \otimes I_{n \times n}$ (Kronecker product). 
I have a function $Y=f(\tilde{\Sigma}^{-1})$ that gives a value in $\mathbb R$. 
Let's define $\Phi_{\Sigma}=vech(\Sigma)$
Now, I am trying to get $\frac{\partial Y}{\partial \Phi_{\Sigma}}$. 
So far I have 
$\frac{\partial Y}{\partial \Phi_{\Sigma}^T} = \Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T \Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$ 
I've been working on this an got that $\Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$ is a vector with $n \times n$ elements. 
Now, working with the first part of the derivative 
$\Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T  = \Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial vec(\Sigma)^T}  D_2 \Bigg)^T  = \Bigg( vec \Big( \frac{\partial \tilde{\Sigma}^{-1}}{\partial \Sigma}\Big)  D_2 \Bigg)^T = \Bigg( vec \Big( \frac{\partial \tilde{\Sigma}^{-1}}{\partial \Sigma^{-1}} \frac{\partial \Sigma^{-1}}{\partial \Sigma} \Big)  D_2 \Bigg)^T$
$= \Bigg( vec \Big( (I_2 \otimes I_n) (-\Sigma^{-1} \Sigma^{-1}) \Big)  D_2 \Bigg)^T$ 
where $D_2$ is the duplication matrix
However, the matrices $(I_2 \otimes I_n)$ and $-\Sigma^{-1} \Sigma^{-1}$ are not conformable. So it is wrong. Also, since  $\Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$ is a vector with $n \times n$ elements, and  $\frac{\partial Y}{\partial \Phi_{\Sigma}}$ is $3 \times 1$, so $\Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T$  should be $3 \times (n \times n)$.
May I ask for advice on solving this task?
 A: For ease of typing, define
$$\eqalign{
&M = \Sigma,\quad 
&N = \Sigma^{-1} \\  
&R = M\otimes I,\quad
&S = N\otimes I = R^{-1},\quad
&f = f(S) \\
&h = {\rm vech}(M),\quad 
&v = {\rm vec}(M) \\ 
&D = D_2,\quad &v = Dh \\
}$$
You don't tell us anything about the function $f(S),\,$ so I'll assume you don't need help
calculating its gradient $G=\left(\frac{\partial f}{\partial S}\right)$
Before we begin, we need a few results from Wikipedia 
and this post which can be summarized
$$\eqalign{
&A\in{\mathbb R}^{m\times n},\quad B\in{\mathbb R}^{p\times q} \\ &I_k\in{\mathbb R}^{k\times k}\qquad \big({\rm Identity\,Matrix}\big) \\
&a = {\rm vec}(A),\quad b={\rm vec}(B)\\
&x={\rm vec}(A^T) = K_{m,n}\,a\quad \big({\rm Commutation\,Matrix}\big) \\
&{\rm vec}(A\otimes B)
 = \left(I_n\otimes K_{q,m}\otimes I_p\right)(I_m\otimes I_n\otimes b)\,a
\\
}$$
Using this, we can write
$$\eqalign{
{\rm vec}(R)
  &= {\rm vec}\big(M\otimes I_n\big) \\
  &= \Big(I_2\otimes K_{n,2}\otimes I_n\Big)
     \Big(I_2\otimes I_2\otimes{\rm vec}(I_n)\Big)\,v \\
  &= Qv \\
}$$
Start by writing the differential of the function, then perform a sequence of changes of variables 
from $S\to R\to v\to h$. 
$$\eqalign{
df
  &= G:dS
\\&= G:(-S\,dR\,S) \\&= -SGS:dR \\
  &= -\operatorname{vec}\left(SGS\right):Q\,dv \\
  &= -Q^T\operatorname{vec}\left(SGS\right):dv \\
  &= -Q^T\operatorname{vec}\left(SGS\right):D\,dh \\
  &= -D^TQ^T\operatorname{vec}\left(SGS\right):dh \\
\frac{\partial f}{\partial h}
  &= -D^TQ^T\operatorname{vec}\left(SGS\right) \\
\\
}$$
The trace/Frobenius product
$\;A:B = {\rm Tr}\big(A^TB\big)\;$
is used in several steps.
The trace's cyclic property allows terms in such products to be rearranged in many ways, e.g.
$$\eqalign{
A:B &= A^T:B^T &= B:A \\
A:BC &= B^TA:C &= AC^T:B \\
}$$
Several steps also made use of the fact that $(M,N)$ and therefore $(R,S)$ are symmetric matrices.
