Degree of the field extension $K(x)\hookrightarrow\operatorname{Quot}\left(K[x,y]/(f)\right) $ > Let $K$ be a field, and $f\in K[x,y]$ an irreducible degree $d$ polynomial. I want to prove that the field extension $$ K(x)\hookrightarrow \operatorname{Quot}\left(K[x,y]/(f)\right) $$ has degree $d$, where $\operatorname{Quot}\left(K[x,y]/(f)\right)$ is the quotient field of the ring $K[x,y]/(f)$. Is it true? I don't know where to start. Any hint would be appreciated.

I've proved in this thread that the field of fractions of $KX,Y$ is $K(X)[Y]/(f)$, so $[K(X)[Y]/(f):K(X)]=\deg_Yf$.