How do you prove that a field is isomorphic to C(x)?

I think that your question is formulated to broad to give a precise answer.

I assume that K is given in the form F(x₁,...,x_k)/(f₁,..,f_m). If your equations f_1...f_m are nice enough (low degre etc.) then several computer algebra packages can tell you transcendence degree of K/F using e.g. Groebner Bases. If this tr. degree is different from one then you are done, otherwise K is the function field of a smooth projective curve C/F.

Now the function field of C/F is isomorphic to F(x) if and only if C is isomorphic to ℙ¹ .

If F is algebraically closed then it suffices to show that C has genus 0. If char(K)=0 this is can be done relatively easy: following the proof of the lemma of the primitive element you can write K=F(x,y)/f. One can consider y as a function on C and this yields a morphism g:C→ ℙ¹. The Riemann-Hurwitz formula (RH) gives you an easy recipe to calculate the genus of C.
(Most of the details are explained in Fulton's book on algebraic curves, if only care about RH you might also read the first two chapters of Silverman's book on Arithmetic of elliptic curves.)
If F is alg. closed, but char(K)>0 this recipe also works, except that the calculation of the entries in the RH formula is harder.
So if F is alg closed then a computer can do the job.

If F is not alg. closed you have a harder problem. You need to use the following criterion
a smooth projective curve C is isomorphic to ℙ¹
if and only if
the genus of C is zero and C has a point with coordinates in F.

So besides the genus calculation you need to show that the curve C has at least one point with coordinates in F. This is in general a hard problem, if F is finite this is known and if F is a number field there is a good criterion to check whether C has a point with coordinates or not, for arbitrary F this seems hard.

Examples ℂ[x,y]/(x^2+y^2+1) is isomorphic to ℂ[z], but ℝ[x,y]/(x^2+y^2+1) is not isomorphic to ℝ[z], since the conic x^2+y^2+1 has non ℝ-points.