How to parametrize a cubic plane curve I've got a cubic surface, but I have no idea how to parametrize it.
If there is no general method, maybe there is one for a specific surface — z^2=xy(x+y-1).
I need to define this surface with the following equations: x=f1(u, v), y=f2(u, v), z=f3(u, v). The simpler f1, f2, f3 are, the better. Polynomials or rational functions will do for me.
 A: In this case, z is a function of x and y, so you can set f1 = u, f2 = v, and f3 = uv(u+v-1). There is no strategy for this in general, you just have to be creative.
A: For the case of an explicit relation $$z=f(x,y)$$ the parameterization is immediate:
$$z=f(x,y).$$
If you prefer,
$$\begin{cases}x=u,\\y=v,\\z=f(u,v).\end{cases}$$
A: As $z^2\geq 0,$ the surface has $3$ sheets, one sheet with $x\geq 0,\;y\geq 0,\;x+y-1\geq 0$, one sheet with $x\leq 0,\; y\geq 0,\; x+y-1\leq 0$ and one sheet with $x\geq 0,\; y\leq 0,\; x+y-1\leq 0.$
For the first sheet, we can set 
\begin{eqnarray}
x_1 &=&(1+v^2) u \\
y_1 &=&(1+v^2)(1-u) \\
z_1 &=& v\,(1+v^2)\, \sqrt{u(1-u)} 
\end{eqnarray}
with $u\in [0,1]$ and $v\in\mathbb{R}.$
The second sheet can be derived from the first one by setting
\begin{eqnarray}
x_2 &=& 1-x_1-y_1 \\
y_2 &=& x_1 \\
z_2 &=& z_1
\end{eqnarray}
because then we have
$$
x_2y_2(x_2+y_2-1) = (1-x_1-y_1)x_1(1-x_1-y_1+x_1-1) \\
= (1-x_1-y_1)x_1(-y_1) = x_1y_1(x_1+y_1-1)
$$
Analogously, we can set the following for the third sheet
\begin{eqnarray}
x_3 &=& y_1 \\
y_3 &=& 1-x_1-y_1  \\
z_3 &=& z_1
\end{eqnarray}
Rational solution:
\begin{eqnarray}
x_1 &=& \frac{(1+v^2) (1-u^2)^2}{(1+u^2)^2}  \\
\\
y_1 &=& \frac{(1+v^2) (2u)^2}{(1+u^2)^2}  \\
\\
z_1 &=& \frac{v\,(1+v^2)\, (1-u^2)(2u)}{(1+u^2)^2} 
\end{eqnarray}
with $u\in [0,1]$ and $v\in\mathbb{R}.$
A: Monge's form with another individual parametrization:
$$ x= u+\frac12, y=v+\frac12, z= \pm \sqrt {(u+\frac12)(v+\frac12)(u+v)};$$
