If $[1,0,0,],[0,1,0],[0,0,1],[1,1,1],[1,3,2],[1,4,3]$ are six points on $P^2$ in general position and $f_0, f_1, f_2, f_3$ are the generators of the four dimensional vector space generated by cubics through the six points as :
$f_0= x_0[-16(x_0)(x_1) + 18(x_0)(x_2) + 4(x_1)^2 - 6(x_2)^2]$,
$f_1= x_0[-6(x_0)(x_1) + 7(x_0)(x_2) - 5(x_2)^2 + 4(x_1)(x_2)]$,
$f_2= - 6(x_0)^2(x_1) + 31(x_0)^2(x_2) - 29(x_0)(x_2)^2 + 4(x_1)^2(x_2)$,
$f_3= 6(x_0)^2(x_1) + 13(x_0)^2(x_2) - 23(x_0)(x_2)^2 + 4(x_1)(x_2)^2$
Define the map $P^2\setminus\{P1,...,P6\}\to P^3$ by $P\to[f_0(p),f_1(p),f_2(p),f_3(p)]$
the image of this map stays on a cubic surface of $P^3$. So, how can get the equation of cubic surface of this image?
any help is appreciate, Thanks
A cubic in $P^3$ has $20$ monomials, so it is uniquely defined by $19$ points in general position. So take $19$ points in general position, plug them into the polynomials, compute all cubic monomials and build a matrix from that. Then determine the kernel of said matrix, which should be one-dimensional. To make things easy, randomly generate a good measure of additional points so you won't have to worry about general position. The resulting matrix should have rank $19$, then it's fine.
Using this approach (and $30$ randomly generated points) I found the resulting cubic to be
\begin{align*} 6f_0^2f_1 - 23f_0f_1^2 + 70f_1^3 - 32f_1^2f_2 + 4f_1f_2^2 + 6f_0^2f_3\\ - 14f_0f_1f_3 + 36f_1^2f_3 - 4f_0f_2f_3 + 5f_0f_3^2 - 6f_1f_3^2&=0 \end{align*}