Show that $\frac{\partial}{\partial X}X^{-1} = \left(-X^{-T} \otimes X^{-1}\right)$ Dear Matrix Calculus experts,

Show that $$\frac{\partial}{\partial X}X^{-1} = \left(-X^{-T} \otimes X^{-1}\right)$$ where $\otimes$ is the Kronecker product and $X$ is a square matrix.


My attempt. Do you experts agree? Many thanks in advance.
\begin{align}
I &= X^{-1} X \\
\Rightarrow 0 &= dX^{-1} X + X^{-1} dX \\
dX^{-1}  &= -X^{-1} dX X^{-1}
\end{align}
Now vectorize both sides, i.e.,
\begin{align}
{\rm vec}\left(dX^{-1}\right)  
&= {\rm vec}\left(-X^{-1} dX X^{-1}\right)\\
&= \left(-X^{-T} \otimes X^{-1}\right) {\rm vec}\left(dX\right) \\
\Rightarrow \underbrace{\frac{\partial}{\partial {\rm vec}(dX)}{\rm vec}\left(dX^{-1}\right)}_{= \frac{\partial}{\partial X}X^{-1} ?} &= \left(-X^{-T} \otimes X^{-1}\right).
\end{align}
 A: Allow me state it generally:
If $dF(X)$ can be expressed as $dF(X)=A(dX)B$, where $F(X)\in\mathbb{R}^{p\times q}$ is a matrix function of $X\in\mathbb{R}^{m\times n}$, then $J_XF(X)=B^T\otimes A$, where the Jacobian matrix $J_XF(X)$ is defined as
\begin{equation}
J_XF(X)\equiv \frac{\partial vec(F(X))}{\partial (vec X)^T}
\end{equation}
Note the transpose sign on the denominator. 
By this rule, given your manipulation  $dX^{-1} = -X^{-1} (dX) X^{-1}$, we can directly confirm the result that you want to show is correct.
So here we focus on the rule itself. 
First vectorize $dF(X)=A(dX)B$ on both sides.
\begin{equation}
d(vec F(X))=vec(A(dX)B)=(B^T\otimes A)vec(dX)=(B^T\otimes A)d(vecX)
\end{equation}
Second we show that 
\begin{equation}
d(vec F(X))=\left[\frac{\partial vec(F(X))}{\partial (vec X)^T}\right] d(vecX) \equiv J_XF(X) d(vecX)
\end{equation}
such that $J_XF(X)=B^T\otimes A$, and the rule valids.
To see this, we write $d(vecF(X))$ explicitly as follows:
\begin{align}
d(vecF(X))=vec(dF(X)) &=
\begin{bmatrix}
df(X)_{11} \\
\vdots \\
df(X)_{p1} \\
\vdots \\
df(X)_{1q} \\
\vdots \\
df(X)_{pq}
\end{bmatrix}
\\
&=
\begin{bmatrix}
\begin{bmatrix}
\frac{\partial f(X)_{11}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{mn}}
\end{bmatrix}
\begin{bmatrix}
dx_{11} \\
\vdots  \\
dx_{m1} \\
\vdots  \\
dx_{1n} \\
\vdots  \\
dx_{mn}
\end{bmatrix}
\\
\vdots
\\
\begin{bmatrix}
\frac{\partial f(X)_{p1}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{mn}}
\end{bmatrix}
\begin{bmatrix}
dx_{11} \\
\vdots  \\
dx_{m1} \\
\vdots  \\
dx_{1n} \\
\vdots  \\
dx_{mn}
\end{bmatrix}
\\
\vdots
\\
\begin{bmatrix}
\frac{\partial f(X)_{1q}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{mn}}
\end{bmatrix}
\begin{bmatrix}
dx_{11} \\
\vdots  \\
dx_{m1} \\
\vdots  \\
dx_{1n} \\
\vdots  \\
dx_{mn}
\end{bmatrix}
\\
\vdots
\\
\begin{bmatrix}
\frac{\partial f(X)_{pq}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{mn}}
\end{bmatrix}
\begin{bmatrix}
dx_{11} \\
\vdots  \\
dx_{m1} \\
\vdots  \\
dx_{1n} \\
\vdots  \\
dx_{mn}
\end{bmatrix}
\end{bmatrix}
\\
&=
\begin{bmatrix}
\frac{\partial f(X)_{11}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{11}}{\partial x_{mn}}
\\
\vdots \\
\frac{\partial f(X)_{p1}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{p1}}{\partial x_{mn}}
\\
\vdots \\
\frac{\partial f(X)_{1q}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{1q}}{\partial x_{mn}}
\\
\vdots \\
\frac{\partial f(X)_{pq}}{\partial x_{11}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{m1}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{1n}} \cdots
\frac{\partial f(X)_{pq}}{\partial x_{mn}}
\end{bmatrix}
\begin{bmatrix}
dx_{11} \\
\vdots  \\
dx_{m1} \\
\vdots  \\
dx_{1n} \\
\vdots  \\
dx_{mn}
\end{bmatrix}
\\
&=
\frac{\partial vec(F(X))}{\partial (vec X)^T} vec(dX)
\end{align}
where $df(X)_{kl}$ is the $k$th row, $l$th col element of $dF(X)$, and it is defined as below, the similar fashion of $dF(X)$.
\begin{equation}
df(X)\equiv\frac{\partial f(X)}{\partial (vecX)^T}d(vecX)
\end{equation}
This completes the proof of the rule.
====================
To clarify the definition of the matrix calculus/differentials:
In some textbooks, we use $J_XF(X)$ or $D_XF(X)$ to express the so-called $\partial F(X)/\partial X$. In wikipedia, it corresponds the "numerator layout". While in other materials we use the gradient notation, namely $\nabla_XF(X)$, or the "denominator layout". $\nabla_XF(X)= [J_XF(X)]^T$.
\begin{equation}
\nabla_XF(X)\equiv \frac{\partial vec^T F(X)}{\partial vec X}
\end{equation}
Hence, it is entire possible that the results in different textbooks varied by transpose. Note pure $\partial F(X)/\partial X$ is not well-defined. 
