Here is an approach that looks different from Jeremy Daniel's one, so may be worth writing:
The idea is simply that $y$ being transcendental over $K$, it acts as a free variable, and by Gauss Lemma, reducibility in $K(y)[t]$ is equivalent to reducibility in $K[y][t]$, the polynomial ring in variables $y,t$.
Let $y := f(x)/g(x) \in K(x)$, with not both $f,g$ in $K$, so that $y\notin K$.
Let $p(t) := f(t) - yg(t) \in K(y)[t]$.
$x$ is transcendental on $K$:
By construction.
$x$ is algebraic on $K(y)$:
Since $p(x) = 0$.
$y$ is transcendental on $K$:
By multiplicativity of degrees of extensions:
$$
\underbrace{[K(x):K]}_{\infty} = \underbrace{[K(x):K(y)]}_{<\infty}[K(y):K].
$$
The polynomial $p(t)$ has degree $n:=\max(\deg f,\deg g)$:
Since the coefficients of $f$ and $g$ don't cancel out (if $f(t) =\sum_i a_i t^i$ and $g(t) = \sum_i b_i t^i$, then $f(t) + yg(t) = \sum_i (a_i + yb_i)t^i$, and $y\notin K$ implies that $a_i + yb_i =0$ iff both $a_i,b_i=0$).
The goal is to show that $p(t)$ is irreducible in $k(y)[t]$, which will imply it is the minimal polynomial for $x$, hence the degree of $x$ is equal to $n$.
$p$ is reducible in $K(y)[t]$ iff it is in $K[y][t]$:
By Gauss Lemma, since $K(y)$ is the field of fractions of $K[y]$.
$K[y][t]$ is the ring of polynomial in two variables over $K$:
Obvious since $y$ is transcendental over $K$.
Now, we consider the polynomial $p(y,t) = f(t) + yg(t) \in K[y,t]$.
If it is reducible, say $h(y,t)k(y,t) = p(y,t)$, then necessarily the "$y$-degree" (=highest occurence of $y$ -- I don't know the usual term) of $h$ is $0$, and that of $k$ is $1$ (wlog), so we may write $h(y,t) = h(t)$ and $k(y,t) = k_1(t) + yk_2(t)$.
Multiplying this yields:
$$
h(t)k_1(t) + yh(t)k_2(t) = f(t) + yg(t),
$$
and by comparing coefficients,
$$
h(t)k_1(t) = f(t),\qquad h(t)k_2(t) = g(t),
$$
hence $h$ divides both $f$ and $g$, contradicting coprimality.
I think this argument should do it, but maybe I am missing something:
For a contradiction, suppose there are two non-zero polynomials $f,g \in k[x]$ such that $(\frac{f}{g})^p=x$. Then $f^p=g^px$. The degree of the polynomial on the left is $\deg(f)*p$ and the degree of the polynomial on the right is $\deg(g)*p+1$. So these are equal. I.e. $$\deg(f)*p=\deg(g)*p+1$$ But this is outrageous and impossible by divisibility considerations.
Best Answer
Since this is an important basic issue, I'll add a complementary answer to Bill Dubuque's, and the good comments above: the definition of $u$ shows that $x$ satisfies a cubic equation over $F(u)$, so is algebraic over $F(u)$. If $u$ were algebraic over $F$, then, by transitivity of "algebraic extension", $x$ itself would be algebraic over $F$.
This less-explicit but more-qualitative kind of argument can succeed when explicit computations become burdensome.