I'm going to decompose Fabian's answer into something a little more newb friendly. An important concept is the partial derivative of a matrix with respect to one of its elements. An example of this is

$$
\partial_{\mathbf{A}_{31}} \mathbf{A}
= \partial_{\mathbf{A}_{31}}
\begin{bmatrix}
\mathbf{A}_{11} & \mathbf{A}_{12} \\
\mathbf{A}_{21} & \mathbf{A}_{22} \\
\mathbf{A}_{31} & \mathbf{A}_{32}
\end{bmatrix}
= \begin{bmatrix}
0 & 0 \\
0 & 0 \\
1 & 0
\end{bmatrix}.
$$

In general, if we take the derivative with respect to the $(i,j)$ entry, then the $(m,n)$ entry of the resulting matrix is

$$
\partial_{A_{ij}} A_{mn} = \delta_{im} \delta_{jn}
$$
where $\delta$ is the Kronecker delta.
This is simply the basic statement of multivariate calculus:
namely $\partial_x x = 1$ and $\partial_x y = 0$.
In particular,

\begin{align*}
\sum_{mn} \partial_{A_{ij}} A_{mn} = \partial_{A_{ij}} A_{ij} = 1.
\qquad \text{(1)}
\end{align*}

To begin the proof, we consider matrix equations, which are

\begin{align*}
\partial_{\mathbf{A}} \text{tr}(\mathbf{A} \mathbf{B} \mathbf{A}^\text{T} \mathbf{C})
&= \partial_{\mathbf{A}} \sum_{m} \left(\mathbf{A} \mathbf{B} \mathbf{A}^\text{T} \mathbf{C}\right)_{mm}
\qquad \text{(Trace[2])} \\
&= \partial_{\mathbf{A}} \sum_{m} \Bigg(\sum_{n k \ell} A_{mn} B_{nk} \left(A^T\right)_{k \ell} C_{\ell m}\Bigg)_{mm}
\qquad \text{ (Matrix multiplication [3])} \\
&= \partial_{\mathbf{A}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m}
\qquad \text{ (Transpose [4])} \\
&= \mathbf{C}^T \mathbf{A B}^T + \mathbf{C A B},
\end{align*}

where we justify the last step component-wise.
For all elements $(i,j)$, it follows that

\begin{align*}
&\Bigg( \partial_{\mathbf{A}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m} \Bigg)_{ij} \\
&= \partial_{\mathbf{A}_{ij}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m}
\qquad \text{(Scalar-by-matrix derivative[5])} \\
&= \sum_{mnk \ell} \partial_{A_{ij}} A_{mn} B_{nk} A_{ \ell k} C_{\ell m}
\qquad \text{ (Linearity of differentiation [6])} \\
&= \sum_{mnk \ell} (\partial_{A_{ij}} A_{mn}) (B_{nk} A_{\ell k} C_{\ell m}) + (A_{mn}) (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m})
\qquad \text{(Product Rule[7])} \\
&= \sum_{mnk \ell} (\partial_{A_{ij}} A_{mn}) B_{nk} A_{\ell k} C_{\ell m}
+ \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m}) \\
&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i}
+ \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m})
\qquad \text{ (Eqn. 1)} \\
&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i}
+ \sum_{mnk \ell} A_{mn} \Big[(\partial_{A_{ij}} A_{\ell k}) (B_{nk} C_{\ell m}) + (A_{\ell k}) (\partial_{A_{ij}} B_{nk} C_{\ell m})\Big]
\text{ (Product R.)} \\
&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i}
+ \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} A_{\ell k}) B_{nk} C_{\ell m} \\
&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i}
+ \sum_{mn} A_{mn} B_{nj} C_{i m}
\qquad \text{ (Eqn. 1)} \\
&= \sum_{k \ell} C_{\ell i} A_{\ell k} B_{jk}
+ \sum_{mn} C_{i m} A_{mn} B_{nj} \\
&= \sum_{k \ell} \left(C^T\right)_{i \ell} A_{\ell k} \left(B^T\right)_{kj}
+ \sum_{mn} C_{im} A_{mn} B_{nj}
\qquad \text{(Transpose)} \\
&= \left( \mathbf{C}^T \mathbf{A} \mathbf{B}^T + \mathbf{C} \mathbf{A} \mathbf{B} \right)_{ij},
\qquad \text{(Matrix multiplication)}
\end{align*}

which completes the proof.