Show that $M = 0$ iff $M_{\mathfrak{p}} = 0$ iff $M_{\mathfrak{m}} = 0$. 
Let $R$ a commutative ring and $M$ an $R$-module. Show that the following statement are equivalent:

*

*$M = 0$,


*$M_{\mathfrak{p}} = 0$ for all $\mathfrak{p} \in \mathrm{Spec}(R)$,


*$M_{\mathfrak{m}} = 0$ for all $\mathfrak{m} \in \mathrm{Specm}(R)$.

Proof :

*

*1) $\implies$ 2): Let $M = 0$. Let $a/b \in M_{\mathfrak{p}}$ (i.e., $b \notin \mathfrak{p}$). Then
$$
  \frac{a}{b} = \frac{a}{1} \cdot \frac{1}{b} = \frac{0}{1}
$$
since $\frac{a}{1} = \frac{0}{1}$ and $\frac{0}{1} \cdot \frac{1}{b} = 0$.


*2) $\implies$ 3): Let $M_{\mathfrak{p}} = 0$ for all $\mathfrak{p} \in \mathrm{Spec}(R)$. Let $\mathfrak{m} \in \mathrm{Specm}(R)$ and $\mathfrak{p} \subset \mathfrak{m}$ a prime ideal. Let $a/b \in M_{\mathfrak{m}}$ (i.e., $b \notin \mathfrak{m}$). Since $b \notin \mathfrak{m}$, then $b \notin \mathfrak{p}$, and thus $a/b \in M_{\mathfrak{p}} = 0$. Therefore $\frac{a}{b} = \frac{0}{1}$.


*3) $\implies$ 1): Suppose $M \neq 0$, i.e., there is $x \in M$ with $x \neq 0$. Let $\mathfrak{m}$ a maximal ideal that contain $x$. Then $\frac{x}{1} = \frac{0}{1}$, and thus $x = 0$ which is a contradiction. Therefore, $M = 0$.
Question: Are my proofs correct ?
 A: $$\boxed{(1) \implies (2)}$$
It seems to be correct, but writing $\frac{a}{b}=\frac{a}{1}\frac{1}{b}$ is a bit unclear somehow. What does $a/1$ really mean? What is the product between $\frac{a}{1}$ and $\frac{1}{b}$?
In order to clarify that, recall that $\dfrac{a}{s}$ is just an equivalence class of $(a,s) \in M \times S$ under some equivalence relation $\sim$.
Proving $ \dfrac{a}{s} = \dfrac{0}{1}$ is just proving that $(a,s) \sim (0,1)$, that is
$$\exists s' \in S,\;  s'(a\cdot 1 - 0 \cdot s) = 0$$
But $a=0$...

$$\boxed{(2) \implies (3)}$$
When you wrote "let $\mathfrak p\subset \mathfrak m$ a prime ideal", actually $\mathfrak p = \mathfrak m$ is fine. As mentioned in the comments, this implication is trivial since any maximal ideal of a commutative unital ring is prime.

$$\boxed{(3) \implies (1)}$$
As mentioned in the comments, saying "Let $\mathfrak m$ a maximal ideal that contain $x$" has no sense, because $x \in M$ while $\mathfrak m \subset R$.
Nevertheless, your idea is the right one (just applied to the wrong object ;-)) : you want some maximal ideal... containing not $x$ (which is not an element of $R$), but rather containing 
$$I := (0 :_R x) = \mathrm{Ann}_R(x) = \{r \in R \mid r \cdot x = 0\}$$
Notice that $x \neq 0 \implies I \neq R$, so you can find a maximal ideal $\mathfrak m$ of $R$ that contains $I$. I let you finish the proof.
