Suppose that the square matrix $M$ with dimensions $(n+m) \times (n+m)$ is partitioned into four submatrix blocks as
$$M_{(n+m) \times (n+m)} = \begin{pmatrix} A_{n \times n } & B_{n \times m} \\ C_{m \times n}&D_{m \times m}\\ \end{pmatrix}.$$
With $A$ and $D$ invertible, we have the Schur complements
$$M/A = D-CA^{-1}B, \quad M/D = A-BD^-1C.$$
Further, partitioned inversion is
$$M^{-1 =} \begin{pmatrix} A & B \\ C&D\\ \end{pmatrix}^{-1}= \begin{pmatrix} A^{-1} + A^{-1} B(M/A)^{-1}CA^{-1} & -A^{-1 }B(M/A)^{-1} \\ -(M/A)^{-1}CA^{-1} & (M/A)^{-1}\end{pmatrix}$$ $$=\begin{pmatrix}(M/D)^{-1} & -(M/D)^{-1}BA^{-1} \\ -D^{-1}C(M/D)^{-1}& D^{-1} + D^{-1} C(M/D)^{-1}BD^{-1}\end{pmatrix}.$$
I can't prove $$ D^{-1} C(M/D)^{-1} = (M/A)^{-1}CA^{-1}.$$
Could you give me some hint?