Find the rational parametrization of this curve I need to find the rational parametrization of the complex curve with affine equation
$$y^2-6x^2y-3x^4+4x^3y+4x^3=0.$$
I did find a solution, I don't know if  there is a more elegant solution, but this is what I did nonetheless.
I took the lines passing through the origin $y=kx$ and intersecting these I find
$$x(k)=\frac{(4-6k)+(4i(k-1))^2\sqrt {k-1}}{2k^2}.$$
Now this is obviously not a rational function, but composing the function $k(t)=t^2+1$ rationalizes the square root and it should give a solution to the problem.Is there a better way to solve this?
 A: Here's a different method using the anticanonical map. The method is very general, but the downside is that some of the intermediate computations are not easily done by hand. See by Rational Parametrizations of Algebraic Curves using a Canonical Divisor by van Hoeij for full algorithmic details.
The basic idea is that we find a birational map $\alpha: C \to Q$ where $Q$ is a plane conic. We are then reduced to finding a parametrization for a conic, which can be done by considering a pencil of lines through a given point. Once we have a parametrization $\varphi: \mathbb{P}^1 \to Q$, we can simply compose with $\alpha^{-1}$ to obtain a parametrization $\newcommand{\P}{\mathbb{P}} \alpha^{-1} \circ \varphi: \P^1 \to C$.
Let $C$ be a curve. In order for there to be rational parametrization for $C$ in the first place, then $C$ must have (geometric) genus $g=0$. In this case, a canonical divisor $K$ of $C$ has degree
$$
\deg(K) = 2g - 2 = -2 \, ,
$$
so its negative, the anticanonical divisor $-K$, has degree $2$. Since $K < 0$, then
$$
\require{cancel}
\ell(-K) - \cancelto{0}{\ell(2K)} = 1 - g + \deg(-K) = 1 - 0 + 2 = 3
$$
by Riemann-Roch. Let $f_1, f_2, f_3$ be a basis for $L(-K)$. (For how to compute such a basis, see the paper cited above, or Computing Riemann-Roch spaces in algebraic function fields and related topics by Florian Hess.) Since $-K$ is very ample, then the induced map
\begin{align*}
\alpha: C &\hookrightarrow \P^2\\
P &\mapsto [f_1(P) : f_2(P) : f_3(P)]
\end{align*}
is a closed embedding. Moreover, since $\deg(-K) = 2$, then its image is a conic $Q$. If $Q$ has a rational point, we can then parametrize it as mentioned above; otherwise we may have to pass to a quadratic extension of the ground field in order to obtain a point.
I used the following Magma code to compute the canonical map for your curve. (You can try it yourself using the online Magma calculator.)
QQ := Rationals();
R<x,y> := PolynomialRing(QQ,2);
f := y^2 - 6*x^2*y - 3*x^4 + 4*x^3*y + 4*x^3;
C0 := Curve(AffineSpace(R),f);
C := ProjectiveClosure(C0);
P<X0,X1,X2> := Ambient(C);
K := CanonicalDivisor(C);
Support(K);
mp := DivisorMap(-K);
P<Y0,Y1,Y2> := Codomain(mp);
mp;

which returned
Mapping from: CrvPln: C to Prj: P
with equations :
X0^3*X2 - 2*X0^2*X2^2 + X0*X1*X2^2
2*X0^3*X2 - 3*X0^2*X2^2 + X1*X2^3
X0^4 - 2*X0^3*X2 + X0^2*X2^2

Running Q := Image(mp); calculates the conic $Q$, which in this case is
$$
Q: Y_0^2 - Y_0 Y_1 + 4 Y_0 Y_2 - 2 Y_1 Y_2 = 0 \, .
$$
Thus we have reduced the problem to parametrizing the conic $Q$. Consider the affine patch where $Y_2 \neq 0$ and let $u = Y_0/Y_2$ and $v = Y_1/Y_2$ on this patch. Then $Q$ is given by
$$
u^2 - uv + 4u  - 2v = 0
$$
on this patch. We see that the point $(0,0)$ lies on $Q$, so we consider the pencil of lines $v = tu$ through this point. Substituting this into the equation above yields
\begin{align*}
0 = u^2 - t u^2  + 4u - 2tu = (1-t) u^2 + (4 - 2t) u = u ((1-t) u + 4 - 2t) \, .
\end{align*}
The root $u = 0$ corresponds to the origin, so we set the other factor equal to $0$ and solve for $u$ in terms of $t$:
$$
0 = (1-t) u + 4 - 2t \implies u = \frac{2t - 4}{1-t} \, .
$$
This allows us to solve for $v$ in terms of $t$ as well:
$$
v = t u = t \frac{2t - 4}{1-t} \, ,
$$
which yields a rational parametrization of the conic $Q$:
\begin{align*}
\psi: \mathbb{A}^1 &\to Q\\
t &\mapsto (u,v) = \left(\frac{2t - 4}{1-t}, t \frac{2t - 4}{1-t}\right) \, .
\end{align*}
With some wrestling with Magma, we can find the inverse of $\alpha$:
KC<xx,yy> := FunctionField(C);
KQ<u,v> := FunctionField(Q);
mpQC := map< Q -> C | [Pushforward(mpCQ, xx), Pushforward(mpCQ, yy), 1]>;
mpQC;

Mapping from: CrvPln: Q to CrvPln: C
with equations : 
-1/4*Y0^2*Y2^2 + 1/2*Y0*Y2^3 - 1/4*Y2^4
1/16*Y0^4 - 3/8*Y0^2*Y2^2 + 1/2*Y0*Y2^3 - 3/16*Y2^4
Y0*Y2^3

Composing these two maps yields the parametrization
\begin{align*}
\varphi: \mathbb{A}^1 &\to C\\
t &\mapsto \left( -\frac{1}{t(t-2)}, \frac{-2t + 1}{t^3 (t-2)} \right) \, .
\end{align*}
You can see from the plots here that this parametrization indeed traces out the curve defined by your implicit equation.
A: Let's work with the closure of the curve and with some variables swapped for practical purposes, the equation is $V(F): xz^2-6xy^2z-3y^4+4y^3z+4xy^3=0$ . The singularities are $(1:0:0),(0:0:1),(1:1:1)$. It can be proved that any quartic with 3 singularities in the points $(1:0:0),(0:1:0),(0:0:1)$ is of the form $g(xy,xz,yz)=0$ where g is the equation of a conic. First of all, we need a projectivity $ω$ to transform our singular points into the three points $(1:0:0),(0:1:0),(0:0:1)$ . One of the projectivities that does the job is the one associated to the matrix $\begin{pmatrix} 1 & -1 & 0 \\ 0 &-1 & 0 \\ 0 &-1 &1 \end{pmatrix}$ . To compute the image of the curve, we substitute $(x:y:z)=(x'-y':-y':z'-y')$ and eliminate the apexes, which gives us the equation $(xy)^2+(xz)^2+(yz)^2-2(xy)(xz)-2(xy)(yz)-2(xz)(yz)=0$  . This substitution is just the polynomial $F∘ω^{−1}$ , the idea is that, if  $G=F∘ω^{-1}$ since $F(ω^{−1}(ω(P)))=F(P)$ then  $G(ω(P))=0⟺F(P)=0$ , so $V(G)= ω(V(F))$ . Now we use the birational mapping  $$φ:\mathbb{P}^{2} \to \mathbb{P}^{2}\\ (x:y:z) \mapsto (xy:xz:yz)=(\frac{1}{z}:\frac{1}{y}:\frac{1}{x})=(X:Y:Z)$$
All that is left to do is, parametrize the conic $X^2+Y^2+Z^2-2XY-2XZ-2YZ=0$ , let's call the rational parametrization $f'$, then $ω^{-1}\circ φ^{-1} \circ f'=f$ is the parametrization for the original curve.
