Part (a) of Exercise 3.4 of Eisenbud's Commutative Algebra 
In the part (a) of Exercise 3.4, it suggests that we may use the relation:
  $${\text{Content}(fg)}\subset{\text{Content}(f)\text{Content}(g)}\subset{\text{rad}\left(\text{Content}(fg)\right)}$$
  to deduce that if $\text{Content}(f)$ contains a nonzerodivisor of $R$, then $f$ is nonzerodivisor of $S=R[x_1,. . . ,x_r]$.

But I don't know how to do this, for example, if all the coefficients of g are in the nilpotent radic of R, assuming that fg=0, the relation above gives nothing. Is there some tips please?
 A: $\DeclareMathOperator{\Ann}{Ann}$
$\DeclareMathOperator{\Ass}{Ass}$
$\DeclareMathOperator{\Content}{Content}$
I don't know if this is the actual proof which Eisenbud had in mind, especially because this proof has nothing to do with the first half of the problem, but here's a possible proof. We start with some preparations:

Lemma Let $R$ be a $\mathbb{Z}^r$-graded ring, and let $M$ be a graded $R$-module. Let $m\in M\setminus\{0\}$ and suppose $P=\Ann(m)$
  is prime. Then $P$ is homogeneous and $P$ is the annihilator of a
  homogeneous element.

The proof of the above lemma is entirely analogous to the one for Proposition 3.12 in Eisenbud, if we endow $\mathbb{Z}^r$ a lexicographic ordering. As a corollary, we obtain:

Corollary Let $R$ be a $\mathbb{Z}^r$-graded ring. If $f\in R$ is a zerodivisor, then there is a nonzero homogeneous element $g\in R$ such that $fg=0$.

(Proof) By hypothesis, there is some nonzero $h\in R$ such that $fh=0$. By replacing $R$ with the subring of $R$ generated by the homogeneous parts of $f$ and $h$, we may assume that $R$ is finitely generated; in particular, $R$ is a Noetherian ring by the Hilbert basis theorem. Since $f$ is a zerodivisor, it is contained in some associated prime $P\in\Ass_{R}(R)$. By the above lemma, $P$ is the annihilator of a homogeneous element. The claim follows. $\blacksquare$

Now back to your question, suppose $f$ is a zero divisor. By the Corollary you can find a nonzero monomial in $S$ which kills $f$, and hence a nonzero element $r\in R$ (hmm, I should have used a symbol other than $R$ in the above corollary; anyhow) which kills $f$. But then $r\Content (f)=0$, so $\Content (f)$ contains no nonzerodivisor.
