Thanks to everyone for the help! However, I am not familiar with any theorems which show that algebraic extensions "behave well in towers." After a little bit more thinking, I have a solution which I think fits better with the methods of my text. I reproduce it here just in case anyone feels like checking it for errors:

*Proof:* Let $\displaystyle v\in K(u)=\{a(u)/b(u):a(x),b(x)\in K[x],b(u)\neq 0\}$ with $\displaystyle v\notin K$, and assume towards a contradiction that $\displaystyle v$ is algebraic over $\displaystyle K$. Then there are $\displaystyle a(x),b(x),c(x)\in K[x]$ with $\displaystyle v=a(u)/b(u)$, and $\displaystyle c(v)=0$. Now, $\displaystyle c(x)$ has the form

$\displaystyle \sum_{i=0}^n c_ix^i$,

where $\displaystyle n=\deg( c(x))$, and each $\displaystyle c_i\in K$. Consider the polynomial

$\displaystyle \sum_{i=0}^n c_i a(x)^i b(x)^{n-i}\in K[x]$. [EDIT: Assuming this polynomial is nonzero,] we have

$\displaystyle \sum_{i=0}^n c_i a(u)^i b(u)^{n-i}$

$\displaystyle =b(u)^n\sum_{i=0}^n c_i (a(u)/b(u))^i$

$\displaystyle =b(u)^n c(a(u)/b(u))$

$\displaystyle =b(u)^nc(v)=0$,

such that $\displaystyle u$ is algebraic over $\displaystyle K$. However, this contradicts the hypothesis that $\displaystyle u$ is transcendental over $\displaystyle K$. So our assumption must be false, and the conclusion follows. $\displaystyle \blacksquare$