<s>[INST] Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$ 
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$

My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
 & \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
 & \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -2y=0\\
 & \iff (3x-y+1)^2-2(x+y)=0.\end{split}
So this look like some crazy parabola, right? But I don't know how to derive parametric equation of this?
 [/INST] I give here another parametrization, a little harder and less practical but somewhat enlightening for the beginners. The focus is $F=\left(\dfrac{-7}{40},\dfrac{11}{40}\right)$ and the directrice is $4x+12y-1=0$ then if $(x,y)$ is any point of the parabola then its distance to the focus and to the directrice are equal by definition of parabola. This gives the equation
$$(40x+7)^2+(40y-11)^2=10(4x+12y-1)^2$$
Because $10=3^2+1$ we have by the known parametrization of solutions of the equation $x^2+y^2=z^2+w^2$
$$\begin{cases}40x+7=rX+sY\\40y-11=rY-sX\\12x+36y-3=rX-sY\\4x+12y-1=rY+sX\end{cases}$$ from which
$$\begin{cases}52x+36y+4=2rX\\4x+52y-12=2rY\\28x-36y+10=2sY\\4x-28y+10=2sX\end{cases}$$ so
$$\frac{13x+9y+1}{x+13y-3}=\frac{2x-14y+5}{14x-18y+5}=t$$ Solving this system we get for $t$ real distinct of $1$
$$x=\frac{-11t^2+86t-59}{200(t-1)^2}$$
$$y=\frac{47t^2-62t+63}{200(t-1)^2}$$
This parametrization is equivalent to the simpler one suggested by the O.P.
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