For the first question alone (without context) I'm going to prove something else first (then check the $\boxed{\textbf{EDIT}}$ for what is asked):
Suppose we have three matrices $A,X,B$ that are $n\times p$, $p\times r$, and $r\times m$ respectively. Any element $w_{ij}$ of their product $W=AXB$ is expressed by:
$$w_{ij}=\sum_{h=1}^r\sum_{t=1}^pa_{it}x_{th}b_{hj}$$
Then we can show that: $$s=\frac {\partial w_{ij}}{\partial x_{dc}}=a_{id}b_{cj}$$
(because all terms, expect the one multiplied by $x_{dc}$, vanish)
One might deduce (in an almost straightforward way) that the matrix $S$ is the Kronecker product of $B^T$ and $A$ so that:$$\frac {\partial AXB}{\partial X}=B^T⊗A$$
Replacing either $A$ or $B$ with the appropriate identity matrix, gives you the derivative you want.
$$\boxed{\textbf{EDIT}}$$
Upon reading the article you added (and after some sleep!), I've noticed that $dD$ is not $\partial D$ in their notation, but rather $\dfrac {\partial f}{\partial D}$ where $f$ is a certain function of $W$ and $X$ while $D=WX$. This means that the first expression you're having problems with is $$\frac{\partial f}{\partial W}=\frac{\partial f}{\partial D}X^T$$
Since the author at the beginning stated that he'd use the incorrect expression "gradient on" something to mean "partial derivative" with respect to that same thing.
So any element of $\partial f/\partial W$ can be written as $\partial f/\partial W_{ij}$. And any element of $D$:
$$D_{ij}=\sum_{k=1}^qW_{ik}X_{kj}$$
We can write $$df=\sum_i\sum_j \frac{\partial f}{\partial D_{ij}}dD_{ij}$$
$$\frac{\partial f}{\partial W_{dc}}=\sum_{i,j} \frac{\partial f}{\partial D_{ij}}\frac{\partial D_{ij}}{\partial W_{dc}}=\sum_j \frac{\partial f}{\partial D_{dj}}\frac{\partial D_{dj}}{\partial W_{dc}}$$
This last equality is true since all terms with $i\neq d$ drop off.
Due to the product $D=WX$, we have $$\frac{\partial D_{dj}}{\partial W_{dc}}=X_{cj}$$ and so $$\frac{\partial f}{\partial W_{dc}}=\sum_j \frac{\partial f}{\partial D_{dj}}X_{cj}$$
$$\frac{\partial f}{\partial W_{dc}}=\sum_j \frac{\partial f}{\partial D_{dj}}X_{jc}^T$$
This means that the matrix $\partial f/\partial W$ is the product of $\partial f/\partial D$ and $X^T$. I believe this is what you're trying to grasp, and what's asked of you in the last paragraph of the screenshot. Also, as the next paragraph after the screenshot hints, you could've started out with small matrices to work this out before noticing the pattern, and generalizing as I attempted to do directly in the above proof. The same reasoning proves the second expression as well...
Best Answer
The differential is correct $$\eqalign{ dF &= dX\,AA^T \cr &= I\,dX\,AA^T\cr }$$ What I normally do at this point is to follow the Magnus-Neudecker convention and apply vec() to both sides $$\eqalign{ {\rm vec}(dF) &= (AA^T\otimes I)\,{\rm vec}(dX) \cr d{\rm vec}(F) &= (AA^T\otimes I)\,\,d{\rm vec}(X) \cr\cr \frac {\partial\,{\rm vec}(F)} {\partial\,{\rm vec}(X)^T} &= AA^T\otimes I \cr }$$ If you don't use vectorization, then you have to deal with $\frac{\partial F}{\partial X}$ as a full-blown fourth-order tensor. In which case index notation is the best way to proceed.
In any case, the derivative is definitely not $AA^T$, which is just a matrix, i.e. a second-order tensor.