when we can say that $ f(x,y)$ and $ g(x,y)$ coprime please?

We say $ f(x,y),g(x,y)$ are coprime, if there does not exist a polynomial $ d(x,y)$ of degree at least one that $ d(x,y)|f(x,y)$ and $ d(x,y)|g(x,y)$.

we take $ f(x,y) = d(x,y)F(x,y),g(x,y) = d(x,y)G(x,y)$ where $ F$ and $ G$ are coprime. we suppsoe that $ \{\begin{array}{c}f(x,y) = d(x,y)F(x,y) = 0\\ g(x,y) = d(x,y)G(x,y) = 0\end{array}$ has finitely many solutions. then $ d(x,y) = 0$ has finitely many solutions. so $ \exists a\in C:\ \forall x\in C: P_{a}(x) = d(x,a)\neq 0$ so $ P_{a}$ is constant then $ \forall (x,y)\in C^{2}:\ d(x,y) = h(y)$ for some $ h\in C[X]$ and like the same we get $ \forall (x,y)\in C^{2}:\ d(x,y) = k(x)$ for some $ k\in C[X]$ so $ d(x,y) = constant$ and $ f$ and $ g$ are coprime the prof is not complete

this is a basic fact in algebraic geometry, usually proven by considering the resultant of the two functions as polynomials in k[x] with coefficients in k[y] or vice versa, as in http://en.wikipedia.org/wiki/Resultant#Applications (it would probably be a bit tricky to come up with this if you hadn't seen it before ?)

In particular, this is a trivial consequence of Bezout's theorem. As such, it doesn't make a great Olympiad problem.

eh, I don't actually know algebraic geometry (especially basic algebraic geometry), but the proof of that theorem usually follows in some way from the technique outlined above, I believe? (so it would be a bit silly to apply bezout).

My point is that Bezout's Theorem should be citable on an Olympiad, which makes this problem a pretty bad idea.

I think not -- my general experience with olympiads is that if you can cite some result that makes the problem utterly trivial, such a solution rarely receives credit and certainly not full credit.