I am reading Theorem 27.5 here which states the following:
Let $f : X \to Y$ be a finite morphism of Noetherian schemes with $Y$ non-singular. Then $X$ is Cohen-Macaulay iff $f_\ast\mathcal{O}_X$ is locally free.
Now implicit in the proof is that if $X$ is Cohen Macaulay, then for any $y \in Y$, $(f_\ast \mathcal{O}_X)_y$ is Cohen-Macaulay. Why is this so? The definition of a Cohen-Macaulay scheme is that $\mathcal{O}_{X,x}$ be Cohen-Macaulay at every $x \in X$ which is not the same as saying that $(f_\ast \mathcal{O}_X)_y $ be Cohen-Macaulay.
A ring $R$ is CM iff $R_{\mathfrak m}$ is CM for every maximal ideal $\mathfrak m$ of $R$, in which case $R_{\mathfrak p}$ is CM for every prime ideal $\mathfrak p$ of $R$. This is, for instance, in Eisenbud (Proposition 18.8). Choose an affine neighbourhood $U$ of $y$ and observe that $\mathcal O_X(f^{-1}(U))$ is CM because its localizations all are. But this means that $(f_\ast\mathcal O_X)(U)$ is CM, hence the localization at $y$ will also be CM.