By inspection, I could find a parameterization for a surface $f(x_1,x_2,x_3)$ in $\mathbb{A}_{C}^3$. Such parameterization is of the form $x_i=\frac{\phi_i(s,t)}{\psi_i(s,t)}$ where $\phi_i(s,t)$ and $\psi_i(s,t)$ are polynomials in the variables $s$ and $t$. My question is: is there any criterion to prove that this parameterization is birational or not? A jacobian, a determinant, something like that?

I tried use MAGMA, via the command IsInvertible. The answer was no: that is not a birational parameterization. However, MAGMA works on $\mathbb{A}_{Q}^3$, and I don't know how to extend the field to complex numbers.

The explicit equation of the surface is,

$f(x_1,x_2,x_3)=2x_1^2x_2 - 2x_3 - 5x_1x_2x_3 + 2x_2x_3^2=0$

One possible parameterization is obtained by solving $f(x_1,x_2,x_3)=0$ for $x_2$, since the variable $x_2$ is linear , and by setting $x_1=s$ and $x_3=t$. Such parameterization is clearly birational, with inverse map $s=x_1$ and $t=x_3$. Explicitly,

$x_1 = s$

$x_2 = \frac{2t}{2s^2-5st+2t^2}$

$x_3 = t$

Another possible parameterization is given by,

$x_1 = \frac{s(-1 + 4t^2)}{3t}$

$x_2 = \frac{2(-1 + t^2)}{3st}$

$x_3 = \frac{2s(-1 + t^2)}{3t}$

In this case, I was not able to find the inverse map, and so I believe that this is not a birational parameterization. But how to prove it?

Thanks in advance,