Finding the equations of an image of a rational map A map from $\mathbb{P^2} \rightarrow \mathbb{P^3}$ is given by: $$f(x:y:z) = (-xyz:xy(x+y+z):xz(x+y+z):yz(x+y+z))$$
How might one find the equations of it's image?
 A: The points where at least one of $x,y,z$ is zero get mapped to one point. For example
$$f(0:y:z)=(0:0:0:yz(y+z))=(0:0:0:1)$$
Let us look at $x,y,z\neq0$ where the bulk of the action is happening. There
$$\begin{align}f(x:y:z)&=f(u:v:1)\\&=(-1:(u+v+1):(u+v+1)/v:(u+v+1)/u)\end{align}$$
We consider $$\begin{align}I&=(u+v+1-X,u+v+1-vY,u+v+1-uZ)\\&=(XYZ-XY-XZ-YZ,vZ-XZ+X+Z,vY-X,u+v-X+1)\end{align}$$

The code in Singular

ring R=0,(u,v,X,Y,Z),(dp(2),lp);
ideal I=(u+v+1-X,u+v+1-v*Y,u+v+1-u*Z);
I=groebner(I);
I;

computes it. The monomial order here compares the part formed by $u,v$ first, only if that is equal it compares the $X,Y,Z$ part of the monomial.

The computation by hand is not too bad, but I never trust my hand with something my computer does better. You can take a look at Buchberger's algorithm. It is not complicated (and is useful to learn), the annoying thing can be running it by hand. Alternatively, use the computer, and then check that it works.
OK. So, now we have that $I\cap k[X,Y,Z]=(XYZ-XY-XZ-YZ)$. Now we homogenize to get the equation in $\mathbb{P}^3$. We put $X=-b/a,Y=-c/a,Z=-d/a$ and multiply by $-a^3$. The $-$ signs is because I left a $-1$ up there when I went to the chart in $\mathbb{P}^3$. We get
$$bcd+abc+abd+acd=0\ \ \ \ \text{ as the equation.}$$
