If $R$ is Noetherian local, then $R$ is CM iff $depth(R)=dim(R)$. If $R$ is only Noetherian, then $R$ is CM iff $R_P$ is CM for all prime ideals.
I want to ask, if $R$ is local, does the two definitions coincide?
The same question is true for regular ring since we have the formula $global.dim(R)=\sup(global.dim(R_P))$. Same for Gorenstein ring. But do we have a similar formula for depth?
Here is a somewhat ad hoc proof of the fact that if $R$ is a local CM ring, then $R_P$ is CM for all prime ideals of $R$. Is this what you wanted?
We first prove that if $P$ is a prime ideal associated to $R$, the height $P=0$. If not, then $\dim R>0$ and the proof is by induction on the depth of $R=\dim R$. So, we have a non-zero element $a\in R$ such that $Pa=0$. Fix a non-zero divisor $x\in R$. Then, we may assume $a$ is not a multiple of $x$, since $\cap (x^n)=0$. If $a=x^nb, a\not\in (x^{n+1})$, then $Pb=0$, since $x\not\in P$. So, $b\not\in (x)$. Then, in $S=R/xR$, $a\neq 0$ and $PSa=0$, so height of $PS$ must be zero by induction ($PS$ is not a prime, but easy to see that we can find a prime ideal containing $PS$ doing what we want). Since $x\not\in P$, height of $P$ itself must be zero. This proves my assertion.
Again, we use the same induction. Let $P$ be any prime. If height of $P=0$, clearly $R_P$ is CM. If the height is positive, from the previous paragraph, we must have a non-zero divisor $x\in P$. Then go modulo $x$ to apply induction and thus the result. Notice that $x$ is a non-zero divisor in $R_P$ too.