You are over-complicating it:
$$\begin{array}{r r c c c l l} \left| \frac1x - \frac16 \right| < \frac6{1000} \implies & -\frac{6}{1000} & < & \frac1x - \frac16 & < & \frac6{1000} \\ & \frac16 - \frac6{1000} & < & \frac1x & < & \frac16 + \frac6{1000} \\ & \frac{964}{6000} & < & \frac1x & < & \frac{1036}{6000} \\ & \frac{6000}{1036} & < & x & < & \frac{6000}{964} \\ & \frac{6216 - 216}{1036} & < & x & < & \frac{5784 + 216}{964} \\ & 6 - \frac{216}{1036} & < & x & < & 6 + \frac{216}{964} \\ & -\frac{216}{1036} & < & x - 6 & < & \frac{216}{964} & \implies \left|x -6\right| < \frac{216}{1036} \end{array}$$
$\displaystyle \begin{align*} \left| \frac{1}{x} - \frac{1}{6} \right| &< \frac{3}{500} \\ \left| \frac{6 - x}{6\,x} \right| &< \frac{3}{500} \\ \frac{\left| 6 - x \right| }{\left| 6\,x \right| } &< \frac{3}{500} \\ \frac{1}{6} \cdot \frac{1}{\left| x \right| } \cdot \left| x - 6 \right| &< \frac{3}{500} \\ \frac{1}{ \left| x \right| } \cdot \left| x - 6 \right| &< \frac{9}{250} \\ \left| x - 6 \right| &< \frac{9}{250} \left| x \right| \end{align*}$
Now if we restrict $\displaystyle \begin{align*} \left| x - 6 \right| < 1 \end{align*}$ say, then that means
$\displaystyle \begin{align*} \left| x \right| - \left| 6 \right| \leq \left| x - 6 \right| &< 1 \textrm{ by the Reverse Triangle Inequality } \\ \left| x \right| - 6 &< 1 \\ \left| x \right| &< 7 \\ \frac{9}{250} \left| x \right| &< \frac{9}{250} \cdot 7 \\ \frac{9}{250} \left| x \right| &< \frac{63}{250} \end{align*}$
So since $\displaystyle \begin{align*} \left| x - 6 \right| < \frac{9}{250} \left| x \right| < \frac{63}{250} \end{align*}$, that means we can set $\displaystyle \begin{align*} \delta = \textrm{min}\,\left\{ 1 , \frac{63}{250} \right\} \end{align*}$.
It's not?
Seems to be EXACTLY what was being asked.i know that abs(f(x)-L) = abs(1/x-1/6) = abs(6-x / 6x) = abs(6-x) abs(1/6x)
But how can i simply it further to match abs(x-6) and what i should bound abs(1/6x) to?
As for how the delta can be written more easily, that is true, but I wanted to leave it in the form that one would have to leave it in more general epsilon-delta proofs.