Generalizing BCnrd's example, almost all surfaces S of degree d ≥ 4 in P^3 have Picard group = Z, hence every curve on a general S is a complete intersection with another surface. Thus any two curves on S intersect. This is the theorem of Noether-Lefschetz.
http://www.springerlink.com/content/t754510m417u0712/
I decided to have a go at finding a parametrisation by following the instructions of Coray and Tsfasman (reference in my comment above), using Magma. Amazingly enough, it works, even working generically with $a,b$ variables.
Here's what I did. Working over the field $K(\omega, \sqrt[3]{a^2b})$, find the three singular points of $X$. Writing $c = \sqrt[3]{a^2b}$, they are $(0:c:a/c:1)$, $(0:\omega^2 c:\omega a/c:1)$, $(0:\omega c:\omega^2 a/c:1)$. There's an obvious rational point $(1:1:0:0)$. The Cremona transformation $f \colon \mathbb{P}^3 \to \mathbb{P}^3$ associated to these four points can be found by linearly mapping them to the standard basis points, applying the standard Cremona transformation $(1/X_0: 1/X_1: 1/X_2: 1/X_3)$, and reversing the linear map. It turns out that $f^{-1}(X)$ is (modulo some rubbish supported on the four planes where $f$ is not defined) a quadric surface $Y$, defined by
$$
X_0^2 + \frac{1}{3a} X_0 X_2 + \frac{1}{27a^2} X_2^2 + \frac{1}{27a^2b}(X_0-X_1)X_3.
$$
The surface $Y$ has an obvious rational point $(0:1:0:0)$, so projecting away from that gives an isomorphism $Y \to \mathbb{P}^2$. The inverse of that isomorphism, composed with $f$, gives a rational map from $\mathbb{P}^2$ to $X$. The equations are (cut & pasted from Magma):
729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 +
12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -3*c^3/a*x^2*y*z^3 +
-c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6
729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 +
12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -9*c^3*x^3*z^3 +
-6*c^3/a*x^2*y*z^3 + -c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6
243*a*c^3*x^5*z + 162*c^3*x^4*y*z + 45*c^3/a*x^3*y^2*z + 6*c^3/a^2*x^2*y^3*z +
1/3*c^3/a^3*x*y^4*z + -1/3*x*y*z^4
-9*c^3/a*x^3*y*z^2 + -3*c^3/a^2*x^2*y^2*z^2 + -1/3*c^3/a^3*x*y^3*z^2 + 1/3*x*z^5
I'm sure these can be tidied up a lot, but notice that they are at least defined over the original base field, since $c$ only ever appears as $c^3$.
I'll post the Magma code (only 20 lines) if anybody's interested.
Best Answer
When I was a student I liked a lot the book by Egbert Brieskorn and Horst Knörrer: Plane Algebraic Curves. I can recommend it.