How to prove a function is not onto? Let $f : Z\to Z$ be the function defined by $f(x) = 3x + 1$. Prove that $f $ is not onto, using a proof by contradiction. (Choose an integer $n$, and then prove ($\forall m \in Z$)($f(m) ≠ n$) by contradiction.)

so far I have:
\begin{gather*}
    1.\quad \text{Let}\quad y  = -1     \qquad   \text{assumption}\\
    2. \quad \text{Let}\quad f(x) = -1       \qquad \text{hypothesis}\\
    3. \quad 3x+1 = -1        \qquad \text{Definition of $f (2)$}\\
    4. \quad 3x  = -2            \qquad \text{Algebra}\\
    5. \quad x = -2/3          \qquad \text{Algebra}
\end{gather*}

What else is needed? I don't know where to go from here?
Your work shows that the only real number $x$ for which $f(x)=−1$ is $x=−2/3$. In particular, there is no integer $n$ such that $f(n)=−1$. Since the domain of $f$ is $\mathbb Z$, the set of integers, you can conclude that $f$ is not surjective.
Let $y=3x+1$ be any element in $\mathbb Z$, then $x=\frac{y-1}{3}$. So, 
$$f(x) = f\left(\frac{y-1}{3}\right) = 3\frac{y-1}{3} + 1 = y$$
Thus, by Definition of onto function we conclude that given function is onto.