= Solution
Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing <monomial> coefficients shows that precisely the terms of even total degree survive. Thus a <vector space> <basis> is
$$
\boxed{\{x^{2a}y^{2b},\ x^{2a+1}y^{2b+1}:a,b\ge0\}.}
$$
Set $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^aw^b$ and $vu^aw^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a <finitely generated algebra>. It is an <integral domain> as a subring of the <integral domain> $k[x,y]$.
The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ to
$$
A(U,W)+VB(U,W).
$$
Its image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct <monomials> are linearly independent. Thus the image vanishes only when both $A,B$ are zero. Consequently
$$
\boxed{R\cong k[U,V,W]/(UW-V^2),\qquad Y=V(UW-V^2)\subseteq\mathbb A^3.}
$$
This gives the requested closed embedding and the <quadratic cone invariant ring> presentation.
The associated <morphism of algebraic varieties> is explicitly
$$
f:\mathbb A^2\longrightarrow Y,\qquad (x,y)\longmapsto(x^2,xy,y^2).
$$
It is surjective. For $(u,v,w)\in Y$, if $u\ne0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.
Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the <Jacobian criterion> makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore
$$
\boxed{\operatorname{Sing}Y=\{(0,0,0)\}.}
$$
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