Problem statement: Assume $V^*$ and $V^0$ are covariance matrices, and the data matrix $X$ satisfies $X^TV^{*-1}X$ and $X^TV_0^{-1}X$ invertible. Please show that $$ (X^TV^{*-1}X)^{-1}(X^TV^{*-1}V_0V^{*-1}X)(X^TV^{*-1}X)^{-1}-(X^TV_0^{-1}X)^{-1} $$ is positive semi-definite (PSD).
Personal thoughts: I had a idea to convert it to $(X^TV^{*-1}X)^{-1}\left[ X^TV^{*-1}(V_0-X(X^TV_0^{-1}X)^{-1}X^T)V^{*-1}X \right](X^TV^{*-1}X)^{-1}$ and then we just need to show the thing in the middle $V_0-X(X^TV_0^{-1}X)^{-1}X^T$ is PSD. But still, I have no clue to do it. I also checked the Woodbury Identity but there lacks a term to make it really work. Any thought how to solve this (or just the original problem)? Thanks for your time!
Your approach so far is a step in the right direction. Note that $V_0-X(X^TV_0^{-1}X)^{-1}X^T$ can be written as the Schur complement of the upper-left block of the matrix $$ M = \pmatrix{X^TV_0^{-1}X & X^T\\ X & V_0} = \pmatrix{X & 0\\0 & I}^T \pmatrix{V_0^{-1} & I\\I & V_0}\pmatrix{X & 0\\0 & I}. $$ It now suffices to argue that $$ P = \pmatrix{V_0^{-1} & I\\I & V_0} $$ is PSD. To that end, note that $$ \pmatrix{V_0^{-1} & I\\I & V_0} = \pmatrix{V_0^{-1/2}\\V_0^{1/2}}\pmatrix{V_0^{-1/2}\\V_0^{1/2}}^T. $$ Thus, the matrix $P$ is PSD, from which it follows that $M$ is PSD, from which it follows that the Schur complement of the upper-left block of $M$ is PSD.