In general, if A and B are square matrices that depend on $x$, the multiplication rule applies: $\frac{d(A B)}{dx} = \frac{dA}{dx} B + A \frac{dB}{dx} $
Applying this, you result follows directly (but, please don't write $B/C$ instead of $B C^{-1}$)
$\frac{d(A -B C^{-1} D)}{dx} = \frac{dA}{dx} - \frac{dB}{dx} C^{-1} D - B \frac{d \, C^{-1}}{dx}D - B C^{-1} \frac{dD}{dx} $
What remains is to compute $\frac{d \, C^{-1}}{dx}$. But deriving $C C^{-1} = I $ we get
$ \frac{dC}{dx} C^{-1} + C\frac{d \, C^{-1}}{dx } = 0 \; \Rightarrow \; \frac{d \, C^{-1}}{dx} = - C^{-1}\frac{dC}{dx}C^{^-1} $
which (check) is the generalization of $\frac{d (y^{-1})}{dx} = - \frac{1}{x^2}\frac{dy} {dx}$
Finally
$\frac{d(A -B C^{-1} D)}{dx} = \frac{dA}{dx} - \frac{dB}{dx} C^{-1} D + B C^{-1}\frac{d \, C}{dx} C^{-1}D - B C^{-1} \frac{dD}{dx} $
which, fixing notation, coincides with your result.