Let $k$ be a field and $K$ be its extension field.
Suppose $y \in K$ is algebraic over $k(x)$ for some $x \in K$ and $y$ is transcendental over $k$.
Then $x$ is algebraic over $k(y)$. I think one way is the following,
Let $ f_0(x)/g_0(x) + f_1(x)/g_1(x)y + f_2(x)/g_2(x)y^2 + f_n(x)/g_n(x)y^n=0$.
so if we take the numerator after adding we get, $f_0(x)g_1(x)...g_n(x) + f_1(x)g_0(x)...g_n(x)y + f_2(x)g_0(x)...g_n(x)y^2+...+f_n(x)g_0(x)..g_{n-1}(x)y^n=0$
Is there any other cleaner way to prove the above statement?