I come across the following claim:
Let $y\in\mathbb{C}[[x]]$ be an algebraic series, that is, there exist $n\in\mathbb{N}^*$ $A_i(x)\in\mathbb{C}[x]$ for $i=0,...,n$ and $A_n(x)\neq 0$ such that \begin{equation}A_n(x)y^n+A_{n-1}(x)y^{n-1}+...+A_0(x)=0.\tag{1}\end{equation}
Then for $P(x,z),Q(x,z)\neq0\in\mathbb{C}[x,z]$, $\frac{P(x,y)}{Q(x,y)}$ can be written as a polynomial on $y$ of degree at most $n-1$, with coefficients in $\mathbb{C}(x)$.
I try to expand $1/Q(x,y)$ and then simplify using the relation $(1)$ but as there are infinitely many terms I don't know how to proceed.
Thanks for your help!
after discussion with the author of the question (see comments) I am compelled to rephrase my answer avoiding what is unnecessary and to bring more detail to the understanding.
as $T^n+p_{n-1}y^{n-1}+...+A_1T+A_0$ is a minimal polynomial of $y$, then $y^n+...+A_1y+A_0=0$ so $y^n=B_{n-1}y^{n-1}+...+B_0$, and step by step all polynomial $f(x,y)$ is expressed as $f(x,y)=C_{n-1}y^{n-1}+...+C_0$, now if $g(x,y)\not=0$ then $\frac{1}{g(x,y)}$ is algebraic over $K(X)$; let $P(T)=irre(\frac{1}{g(x,y)},K(X))=T^s+p_{s-}+...+p_1T+p_0$ therefore $p_s(\frac{1}{g(x,y)})^s+...+p_1\frac{1}{g(x,y)}+p_0=0$, by multiply this equality by $g(x,y)^{(s-1)}$ we get the expression of $\frac{1}{g(x,y)}$ as a polynomial in $y$ with coefficient in $K(X)$. finally we conclude that all$f(x,y)$ and $g(x,y)$ with $g(x,y)\not=0$ we get $\frac{f(x,y)}{g(x,y)}$ as a polynomial in $K(x)[y]$ at most of degree $n-1$.