Let $k$ be an algebraically closed field and define $f : k^3 \longrightarrow k^3$ by $$f(x,y,z) = (x, xy, xyz).$$
I would like to verify that the image of this map is $$f(k^3) = \{ (0,0,0) \} \cup k^3 \backslash \{ x,y =0 \},$$ where $f^{-1}(k^3 \backslash \{ x,y =0 \}) = k^3 \backslash \{ x,y =0 \}$ and $f^{-1}(\{ (0,0,0) \}) = \{ (0,0,z) : z \in k \}$.
Notice that (if my computation is correct) the image is neither open or closed in $k^3$. How does one determine if the image is dense?
Please note that $k^3 = \mathbb{A}_k^3$ equipped with the Zariski topology.
Context: qualifying exam preparation.
Let $(a,b,c)$ in $f(k^3)$, then , ther exists $(x,y,z)\in k^3$ such that:
$$a=x,b=xy,c=xyz$$ If $a\neq 0$ and $b \neq 0$ then: $$x=a, y=\frac ba, z=\frac cb$$
If $a=0$ then $x=0$ and $b=0$ and $c=0$ . Thus $a=b=c=0$
If $b=0$ then $x=0$ or $y=0$ Thus: $b=c=0$
That shows that (whre $k^*=k\backslash\{0\}$): $$f(k^3)=(k^*\times k^*\times k) \cup (k\times\{0\}\times\{0\}) $$ and forall $(a,b,c)\in k^3$, we have :
$f(a,0,0)=(a,0,0)$ and if $ab\neq 0$, then: $f(a,\frac ba, \frac cb)=(a,b,c)$