Regarding element-wise derivative of matrices 
Let $X=(x_{ij})$ be a real n-by-n matrix where $x_{ij}$ are in a range such that $X$ is insvertible. Let $Y=X^{-1}=(y_{pq})$, and regard  $y_{pq}=y_{pq} (x_{11},x_{12},\ldots,x_{nn})$ as a function of $x_{ij}$. Prove that $\frac{\partial y_{pq}}{\partial . x_{ij}}=-y_{pi}y_{jq}.$.

Any solution to prove this in a right way would be appreciated. And is there any name for this type of derivative? How can I handle it?
And the followings are some non-sense scribbles I've made
$XY=I$
${d(XY)\over dx}=IY+X{dY\over dx}=O$
${\partial Y\over \partial x_{ij}}=-X^{-1}Y=-Y^2$
$\frac{\partial y_{pq}}{\partial x_{ij}}=-y_{pi}y_{jq}.$.
Is there any why this can be justified?
 A: I guess the argument is cleaner if you use:
$$
0=\frac{\partial (XY)}{\partial x_{ij}} =\frac{\partial X}{\partial x_{ij}}Y+ X\frac{\partial Y}{\partial x_{ij}}
$$
So
$$
\frac{\partial Y}{\partial x_{ij}}=-X^{-1}\frac{\partial X}{\partial x_{ij}}Y=-Y\frac{\partial X}{\partial x_{ij}}Y
$$
So taking the $(l,k)$ matrix entry in the previous equation we get:
$$
\frac{\partial y_{lk}}{\partial x_{ij}}=-\sum_{m,n} y_{ln}\left(\frac{\partial X}{\partial x_{ij}}\right)_{nm}y_{mk}
$$
And $\left(\frac{\partial X}{\partial x_{ij}}\right)_{nm}=\delta_{in}\delta_{jm}$ so:
$$
\frac{\partial y_{lk}}{\partial x_{ij}}=-\sum_{m,n} y_{ln}\delta_{in}\delta_{jm}y_{mk}=-y_{li}y_{jk}
$$
A: $\frac{\partial X}{\partial x_{ij}} = \frac{\partial x_{pq}}{\partial x_{ij}} = \delta_{pi}\delta_{jq}$ is a 4th order tensor.
$I = \delta_{pq}$ is a 2nd order tensor (or a matrix).
So be careful, $\frac{\partial X}{\partial x_{ij}} \neq I$
$XY=I \Leftrightarrow x_{pk}y_{kq} = \delta_{pq}$
$\frac{\partial (x_{pk}y_{kq})}{\partial x_{ij}} = O_{pqij}$, where O is the null 4th order tensor.
$\frac{\partial x_{pk}}{\partial x_{ij}}y_{kq} + x_{pk}\frac{\partial y_{kq}}{\partial x_{ij}} = O_{pqij}$
$x_{pk}\frac{\partial y_{kq}}{\partial x_{ij}} = -\frac{\partial x_{pk}}{\partial x_{ij}}y_{kq} = -\delta_{pi}\delta_{jk}y_{kq} = -\delta_{pi}y_{jq}$
$y_{pt}x_{tk}\frac{\partial y_{kq}}{\partial x_{ij}} = -y_{pt}\delta_{ti}y_{jq}$
$\delta_{pk}\frac{\partial y_{kq}}{\partial x_{ij}} = -y_{pi}y_{jq}$

$\frac{\partial y_{pq}}{\partial x_{ij}} = -y_{pi}y_{jq}$

