By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$ 
By using the definition of a limit only, prove that
$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$

We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$
I have simplified $\left|\dfrac{1}{3x+1}-1\right|$ down to $\left|\dfrac{-3x}{3x+1}\right|$
Then since ${x\rightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right|<\left|\dfrac{-3x}{4}\right|<\left|\dfrac{-3\delta}{4}\right|=\epsilon$$
No sure if the solution is correct

No. If we assume $-1 < x < 1$, then  $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $\left|\dfrac {1}{3x+1}\right|$ will be unbounded.
If you want to put an upper bound on $\left|\dfrac {1}{3x+1}\right|$, then you are going to need something like  $3x + 1 > \frac 13$, which is equivalent to 
$x > -\frac 29$.
So, $|x| < \frac 29 \implies 3x+1 > \frac 13 \implies
\left|\dfrac {1}{3x+1}\right|<3$.
So, then, $\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right|
< |9x|$ 
To make that less than $\epsilon$, we need to have $|x|<\frac 19 \epsilon$.
So we need to have $\delta < \frac 19\epsilon$ and $\delta < \frac 29$.
Hence we define $\delta = \min(\frac 19\epsilon, \frac 29)$
