How to solve this system of non-linear equations.

Question

$$\begin{cases}2x-y^2=0\\ 4y^3-2yx-3z=0\\ 3z^2-3y=0 \end{cases}$$

I know the solutions are $(0,0,0)$ and $(\frac{1}{2},1,1)$ but I just don't know how to manipulate the equations to obtain them.

Answer 1

from the first equation we get $$x=\frac{y^2}{2}$$ plugging this in the second equation we get $$4y^3-y^3=z$$ or $$3y^3=z$$ plugging this in the last equation we obtain $$y^6-y=0$$ factorizing $$y(y^5-1)=0$$ therefor $$y=0$$ or $$y=1$$ can you finish from here?