Let $\mathbb{C}$ be the complex field and $\mathbb{C} [x, y, z]$ be a three-variable polynomial ring.
$A = \mathbb{C} [x, y, z] / (y ^ 2z-x ^ 3 + 2z ^ 3)$ is an integral domain.
Let the image of $f \in \mathbb{C} [x, y, z]$ in $A$ be $\overline{f}$. Let $ Q (A) $ be the field of fractions of $ A $.
$K = \{\overline{f} / \overline{g} \mid \text{$f, g$ are homogeneous polynomials and $\deg (f) = \deg (g)$}\}\cup\{0\}$
is a subfield of $Q (A)$.
I think $K$ is a function field of an elliptic curve defined by $y ^ 2 = x ^ 3-2$. Therefore, the transcendence degree of $K$ over $\mathbb{C}$ should be $1$.
Please show that the transcendence degree over $\mathbb{C}$ of $K$ is $1$.
Question: "Please show that the transcendence degree on $C$ of $K$ is 1."
Answer: The map $\phi$ has the following property: let $f(u,v)\in k[u,v]$ with $f(u,v):=\sum_{i+j \leq d}\alpha_{(i,j)}v^iu^j$. It follows
$$\phi(f):= \sum_{i,j} \alpha_{(i,j)}(x/z)^i(y/z)^j =$$
$$ \frac{1}{z^d}\sum_{i,j} \alpha_{(i,j)}x^iy^j z^{d-(i+j)}= \frac{F(x,y,z)}{z^d}$$
where $F$ is a homogeneous polynomial of degree $d$. Hence $\phi(f) \in K$ is in your subring $K \subseteq Q(A)$. Since $Q(B)$ is a field and $\phi$ is a map of rings, you get an injective map of rings
$$\phi: Q(B) \rightarrow K$$
and it remains to prove that this map is surjective. An element in $K$ is on the form $\frac{F(x,y,z)}{G(x,y,z)}$ where $F,G$ are homogeneous polynomials of the same degree $d$ and you may write $F=z^df(x/z,y/z)$ where $f(u,v)$ is a polynomial of degree $\leq d$ and similar for $G=z^dg(x/z,y/z)$. Hence
$$\frac{F}{G}=\frac{z^df(x/z,y/z)}{z^dg(x/z,y/z)} \cong \frac{f(x/z,y/z)}{g(x/z,y/z)}.$$
It follows
$$\phi(\frac{f(v,u)}{g(v,u)})=\frac{F}{G}$$
and the map $\phi$ is surjective.