Add $ \dfrac{-1}{-3z^3+3z^4} $ and $ \dfrac{1}{-3+3z} $ to get $ \dfrac{ \color{purple}{ z^3-1 } }{ 3z^4-3z^3 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{z^3}$.

$$ \begin{aligned} \frac{-1}{-3z^3+3z^4} + \frac{1}{-3+3z} & = \frac{ -1 }{ -3z^3+3z^4 } + \frac{ 1 \cdot \color{blue}{ z^3 }}{ \left( -3+3z \right) \cdot \color{blue}{ z^3 }} = \\[1ex] &=\frac{ \color{purple}{ -1 } }{ -3z^3+3z^4 } + \frac{ \color{purple}{ z^3 } }{ -3z^3+3z^4 }=\frac{ \color{purple}{ z^3-1 } }{ 3z^4-3z^3 } \end{aligned} $$

Simplify $ \dfrac{z^3-1}{3z^4-3z^3} $ to $ \dfrac{z^2+z+1}{3z^3} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{z-1}$.

$$ \begin{aligned} \frac{z^3-1}{3z^4-3z^3} & =\frac{ \left( z^2+z+1 \right) \cdot \color{blue}{ \left( z-1 \right) }}{ 3z^3 \cdot \color{blue}{ \left( z-1 \right) }} = \\[1ex] &= \frac{z^2+z+1}{3z^3} \end{aligned} $$