Block multiplication

Setup. Let $M^{-1} = \begin{pmatrix} X & Y \\ 0 & Z \end{pmatrix}$ (the lower-left block must be zero because the product with the lower-left block of $M$ is zero and the upper blocks are invertible). Then
$$MM^{-1} = \begin{pmatrix} AX+0 & AY+BZ \\ 0 & CZ \end{pmatrix} = \begin{pmatrix} I & 0 \\ 0 & I \end{pmatrix}.$$

Solving. From $CZ = I$, we get $Z = C^{-1}$. From $AX = I$, we get $X = A^{-1}$. From $AY + BZ = 0$, we get $Y = -A^{-1}BZ = -A^{-1}BC^{-1}$.

Conclusion.
$$M^{-1} = \begin{pmatrix} A^{-1} & -A^{-1}BC^{-1} \\ 0 & C^{-1} \end{pmatrix}.$$