$M$ is a finitely generated $A$-module of dimension $d$ such that $G(M)$ is equidimensional and $M$ does not have any embedded prime.
Given $x\in I$, where $I$ is an ideal of $A$, and $\dim G(M)/x^*G(M)<d$ then show that $x$ is non zero divisor of $M$.
Not everybody has the book. So it would be nice if you can explain your notation. Perhaps read how to ask question article.
I believe that your setting is the following: $(R,m)$ is a Noetherian local ring. Let $I \subseteq m$ be an ideal and $M$ a finitely generated $R$-module. Let $G(-)$ denote $gr_I(-)$, and let $x^*$ denote the initial form of $x$ in $G(-)$.
Question: Let $d = \dim R$. Assume that $M$ is failthful and unmixed and $G(M)$ is equidimensional. Let $x \in I$. If $\dim G(M)/x^* G(M) < d$, then $x$ is a non zero-divisor on $M$.
Consider the following exact sequence $$ 0 \to L \to G(M) \to G(M/xM) \to 0 $$ where $L$ is the kernel of the map $G(M) \to G(M/xM)$. It is easy to see that $x^* G(M) \subseteq L$. Hence we have $$ 0 \to L / x^* G(M) \to G(M)/ x^* G(M) \to G(M/xM) \to 0 $$ exact. By the hypothesis the dimension of $G(M)/ x^* G(M)$ is at most $d-1$. This implies that $\dim G(M/xM)$ is at most $d-1$. Since $\dim G(M/xM) = \dim M/xM$, you can conclude $\dim M/xM = d-1$. Since $x$ avoids all the minimal primes ideal of $M$ and $M$ unmixed, $x$ is a non zerodivisor of $M$.
I don't think I used the assumption of $G(M)$ being equidimensional.
I have already solved the question. If you check that there is 1-1 correspondence between minimal primes of Rees module of $M$ and minimal primes of $M$, then using equidimensionality of $G(M)$ you will get the answer.
