Proof.
Note that the assumption on Y' \to Y implies that for y' \in Y' mapping to y \in Y the field extension \kappa (y')/\kappa (y) is finitely generated. Hence also all the fibres X'_{y'} = (X_ y)_{\kappa (y')} are locally Noetherian, see Varieties, Lemma 33.11.1. Thus the lemma makes sense. Set y' = f'(x') and y = f(x). Hence we get the following commutative diagram of local rings
\xymatrix{ \mathcal{O}_{X', x'} & \mathcal{O}_{X, x} \ar[l] \\ \mathcal{O}_{Y', y'} \ar[u] & \mathcal{O}_{Y, y} \ar[l] \ar[u] }
where the upper left corner is a localization of the tensor product of the upper right and lower left corners over the lower right corner.
Assume f is Cohen-Macaulay at x. The flatness of \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x} implies the flatness of \mathcal{O}_{Y', y'} \to \mathcal{O}_{X', x'}, see Algebra, Lemma 10.100.1. The fact that \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x} is Cohen-Macaulay implies that \mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'} is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that f' is Cohen-Macaulay at x'.
Assume f is flat at x and f' is Cohen-Macaulay at x'. The fact that \mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'} is Cohen-Macaulay implies that \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x} is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that f is Cohen-Macaulay at x.
Assume Y' \to Y is flat at y' and f' is Cohen-Macaulay at x'. The flatness of \mathcal{O}_{Y', y'} \to \mathcal{O}_{X', x'} and \mathcal{O}_{Y, y} \to \mathcal{O}_{Y', y'} implies the flatness of \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}, see Algebra, Lemma 10.100.1. The fact that \mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'} is Cohen-Macaulay implies that \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x} is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that f is Cohen-Macaulay at x.
\square
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