To get the complex solutions $z=x+yi$ of $(z-i)^3=-8i$ we do the following: $$\left (x+yi-i\right )^3=-8i \Rightarrow \sqrt[3]{\left (x+yi-i\right )^3}=\sqrt[3]{8(-i)} \Rightarrow x+yi-i=2i \Rightarrow x+(y-1)i=2i $$ From that we get $x=0$ and $y-1=2 \Rightarrow y=3$.
Therefore the compelx solutions are $z=3i$ and $z=-3i$.
Is everything correct?
On
No, it is not correct. Your solution $z=-3i$ does not satisfy your equation.
To solve the equation $$ (z-i)^3 = -8i$$ you may let $w=z-i$ and solve $w^3=-8i$ which gives you three solutions for $w$
$w=-2e^{i\pi /6}$, $w=-2e^{i\pi /6 +2 \pi i/3}$,$w=-2e^{i\pi /6 +4\pi i /3}$
Then you find $z=w+i$ for each $w$ and you are done.
Let $w=z-i$, you want to solve for $w^3=-8i$. This means $$w^3+8i=(w)^3-(2i)^3=(w-2i)(w^2+2iw-4)=0$$ So $w=2i$ is one solution. The other two solutions are (by solving the quadratic) $$w=-i+\sqrt{3} \text{ and } w=-i-\sqrt{3}.$$ Thus $$z=3i, \sqrt{3}, -\sqrt{3}$$