Too long for a comment: why $K(X)$ is not $K$-isomorphic to $K(X_1, X_2)$ ?
Take $r=R(X) \in K(X) \backslash K$. It is not hard to see that $X$ satisfies an algebraic equation over $K(r)$. Therefore, the extension
$K(r)\subset K(X)$ is finite ( it is interesting to find the exact degree, but we won't concern with it now). So, if $r_1$, $r_2$ in $k(X) \backslash K$, the extension $K(r_1)\subset K(r_1, r_2) \subset K(X)$ is finite, and so $r_2$ is algebraically over $K(r_1)$. This is equivalent to $r_1$, $r_2$ algebraically dependent over $K$. Now, this will be true also if one of the $r_i$ is in $K$ ( that is even simpler to see).
But $X_1$, $X_2$ from $K(X_1, X_2)$ are not algebraically dependent over $K$.
(We showed that the two extensions have a different transcendent degree over $K$).
As example to the question, you can take $K= \mathbb{Q}$, and $u=v= \pi$, or any other transcendental number.