Calculating the matrix derivative $\frac{\partial}{\partial X}||CX^{-1}B||_F^2$ For matrices $C, X$, and $B$, I know that $\frac{\partial}{\partial X}||CXB||_F^2 = 2C^TCXBB^T$, and that $\partial X^{-1} = -X^{-1}(\partial X) X^{-1}$. However, I am unable to combine these results to calculate $\frac{\partial}{\partial X}||CX^{-1}B||_F^2$.

Define a new matrix variable
$$\eqalign{
 Y &= X^{-1} \qquad\implies\qquad
dY &= -Y\,dX\,Y \\
}$$
Use the gradient, which you already know, to write the differential of the function in terms of $Y$
$$\eqalign{
\phi &= \|CYB\|^2 \\
d\phi &= 2\Big(C^TCYBB^T\Big):dY \\
}$$
Then perform a change of variables from $Y\to X$
$$\eqalign{
d\phi
 &= 2\Big(C^TCYBB^T\Big):\Big(-Y\,dX\,Y\Big) \\
 &= -2\Big(Y^TC^TCYBB^TY^T\Big):dX \\
 &= -2\Big((X^{-1})^TC^TCX^{-1}BB^T(X^{-1})^T\Big):dX \\
\frac{\partial \phi}{\partial X}
 &= -2(X^{-1})^TC^TCX^{-1}BB^T(X^{-1})^T \\
}$$

In the above, a colon is used as a product notation for the trace, i.e.
$$\eqalign{A:B = {\rm Tr}(A^TB) = {\rm Tr}(B^TA) = B:A}$$
The terms in such a product can be rearranged in a number of equivalent ways, e.g.
$$\eqalign{
A:B &= A^T:B^T \\
A:BC &= B^TA:C = AC^T:B \\
}$$
due to the properties of the trace function.
As you have discovered, the chain rule is difficult to apply in Matrix Calculus. It often requires the calculation of intermediate quantities which are third and fourth order tensors.
The beauty of the differential approach is that the differential of a matrix acts like a matrix. In particular, it obeys all of the rules of matrix algebra.
