Answer
$\left\{\dfrac{2-\sqrt{3}}{3},\dfrac{2+\sqrt{3}}{3}\right\}$
Work Step by Step
In the form $x^2+bx=c,$ to complete the square of the expression at the left side, add $\left(\dfrac{b}{2}\right)^2$ to both sides of the equation. Thus, the given equation, $
z^2-\dfrac{4}{3}z=-\dfrac{1}{9}
,$ is equivalent to
\begin{align*}
z^2-\dfrac{4}{3}z+\left(\dfrac{-4/3}{2}\right)^2&=-\dfrac{1}{9}+\left(\dfrac{-4/3}{2}\right)^2
\\\\
z^2-\dfrac{4}{3}z+\left(\dfrac{-2}{3}\right)^2&=-\dfrac{1}{9}+\left(\dfrac{-2}{3}\right)^2
\\\\
z^2-\dfrac{4}{3}z+\dfrac{4}{9}&=-\dfrac{1}{9}+\dfrac{4}{9}
\\\\
z^2-\dfrac{4}{3}z+\dfrac{4}{9}&=\dfrac{3}{9}
.\end{align*}
Using $a^2\pm2ab+b^2=(a\pm b)^2$ or the factoring of perfect square trinomials, the equation above is equivalent to
\begin{align*}
\left(z-\dfrac{2}{3}\right)^2&=\dfrac{3}{9}
.\end{align*}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{align*}
z-\dfrac{2}{3}&=\pm\sqrt{\dfrac{3}{9}}
\\\\
z-\dfrac{2}{3}&=\pm\sqrt{\dfrac{1}{9}\cdot3}
\\\\
z-\dfrac{2}{3}&=\pm\sqrt{\left(\dfrac{1}{3}\right)^2\cdot3}
\\\\
z-\dfrac{2}{3}&=\pm\dfrac{1}{3}\sqrt{3}
\\\\
z-\dfrac{2}{3}&=\pm\dfrac{\sqrt{3}}{3}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
z&=\dfrac{2}{3}\pm\dfrac{\sqrt{3}}{3}
\\\\
z&=\dfrac{2\pm\sqrt{3}}{3}
.\end{align*}
Hence, the solution set of the equation $
z^2-\dfrac{4}{3}z=-\dfrac{1}{9}
$ is $\left\{\dfrac{2-\sqrt{3}}{3},\dfrac{2+\sqrt{3}}{3}\right\}$.