= Solution
Two irreducible <algebraic variety>[varieties] $X$ and $Y$ are <birational variety>[birational] if they contain nonempty <Zariski-open set>[Zariski-open subsets] $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their <function field of an algebraic variety>[function fields] are isomorphic:
$$
k(X)\cong k(Y).
$$
For an irreducible variety,
$$
\dim X=\operatorname{trdeg}_k k(X).
$$
An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the <dimension from the function field>.
Now let $K/\mathbb C$ be a finitely generated field extension. Choose field generators $a_1,\ldots,a_m$ and set
$$
A=\mathbb C[a_1,\ldots,a_m]\subseteq K.
$$
Then $A$ is a finitely generated integral $\mathbb C$-algebra and $\operatorname{Frac}(A)=K$. The affine irreducible variety
$$
X_0=\operatorname{Spec}A
$$
has function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space $\mathbb A^N$, identify that affine space with the standard open chart of <projective space> $\mathbb P^N$, and take the projective closure $X$ of $X_0$. The closure of an <irreducible topological space> is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a <projective variety> and
$$
k(X)=k(X_0)=K.
$$
This constructs the <projective model of a finitely generated field>.
For the curve
$$
X=V\bigl(y^2-x(x-1)^2\bigr),
$$
put
$$
t=\frac y{x-1}.
$$
In its function field, the defining equation gives
$$
t^2=x,
\qquad
y=t(x-1)=t(t^2-1).
$$
Hence $\mathbb C(X)=\mathbb C(t)$, so $X$ is rational. Equivalently, this is the <normalization of y squared equals x times x minus one squared>, whose smooth projective normalization is $\mathbb P^1$ and has <geometric genus> zero.
A <smooth projective curve> in the <projective plane> of degree $d$ has, by the <genus of a smooth plane curve> formula,
$$
g=\frac{(d-1)(d-2)}2.
$$
Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb C$ are rational, so each is birational to $X$. Therefore the required degrees are exactly
$$
\boxed{d=1\text{ or }d=2}.
$$
Finally, in $\mathbb A^3_{x,y,z}$ take
$$
S=V(F),
\qquad
F(x,y,z)=y^2-x(x-1)^2.
$$
The polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1)^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible <affine hypersurface> of dimension two. Its function field is
$$
\mathbb C(S)=\mathbb C(t,z),
\qquad
x=t^2,
\qquad
y=t(t^2-1),
$$
so $S$ is birational to $\mathbb A^2$.
The partial derivatives are
$$
F_x=-(x-1)(3x-1),
\qquad F_y=2y,
\qquad F_z=0.
$$
Solving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the <Zariski tangent space>[Jacobian criterion],
$$
\operatorname{Sing}(S)=\{(1,0,z):z\in\mathbb C\}\cong\mathbb A^1.
$$
This is an irreducible subvariety of dimension one, giving the requested <singular cylinder over a nodal curve>.
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