An exercise states:
Decompose into irreducible components the closed set $ X \subset \mathbb{A}^{3} $ defined by $ y^{2} = xz,\; z^{2} = xy. $ Prove that all its components are birational to $ \mathbb{A}^{1}. $
My first stumbling block is that I'm not sure how to decompose $ V(y^{2}-xz,\; z^{2}-y^{3}) $ beyond recognising(if it is indeed so) that this is the same as $ V(xy-z). $ Beyond that, how can I establish that there is a birational map $ \varphi: V(xy-z) \dashrightarrow \mathbb{A}^{1}$?
There are four irreducible components, let us parametrize them ( I will let you turn it into a rigourous proof ).
If $y=0$ (resp. $z =0$) then you get $z = 0$ (resp. $y=0$). It means that $(x,0,0)$ is a component (indeed isomorphic to $\Bbb A^1$) given by the equations $y=z=0$.
If $y \neq 0$ and $z \neq 0$ then $y^2/z = z^2/y$ i.e $y^3 = z^3$. This is the union of three lines given by $y = j^kz$ where $j^3 =1$ and $j \neq 1$. For each of these lines, you get a component of $X$ given by the equations $y = j^kz, x = j^{2k}y$.
Finally, notice that $V(xy-z)$ is a surface and hence can't be isomorphic to your variety which is a curve.