$\mathbf {The \ Problem \ is}:$ Let $\phi : \mathbb C^3 \to \mathbb C^3$ be the map $(x,y,z) \mapsto (x+y+z,xy+yz+zx,xyz)$, then show that $\phi$ is an open map .
$\mathbf {My \ approach} :$ First of all, note that $\phi$ is onto and a closed map .
Then, let $\phi \equiv (\phi_1,\phi_2,\phi_3)$, then $\phi_1$ is open but obviously $\phi_2$ and $\phi_3$ are also open , and $\phi$ is not injective too as it is symmetric in $x, y$ and $z$ but how to show that $\phi$ is open ???
I can't approach further, a small hint is appreciated .
It is not true that $\phi_2$ and $\phi_3$ are not open. However, it is not clear that this yields anything further.
Let $A=(x, y, z)\in U$ where $U$ is an open subset of $\mathbb C^3$, suppose that there is an ball of radius $r>0$ centered at $A$, which is contained in $U$, and write $B=\phi_2(x, y, z)=xy+yz+xz$. If $y+z\neq 0$ then $\phi_2(U)$ contains an open ball of radius $r(y+z)$ centered at $B$ since we may write $\phi_2(x', y, z)=x'(y+z)+yz=B+(x-x')(y+z)$ and vary $x'$ in an open ball of radius $r$ around $x$. If $y+z=0$ then $\phi_2(U)$ contains an open ball of radius $(r/2)^2$ around $B$ since we may write $\phi_2(x, y+\epsilon, z-\epsilon)=B+\epsilon^2$ and vary $\epsilon$ in a ball of radius $r/2$ around 0.
Let $A=(x, y, z)\in U$ where $U$ is an open subset of $\mathbb C^3$, suppose that there is an ball of radius $r>0$ centered at $A$, which is contained in $U$, and write $B=\phi_3(x, y, z)=xyz$. If $x=y=z=0$ then $\phi_3(U)$ contains a ball of radius $(r/3)^{3}$ around $B$ by observing that $\phi_3(x+\epsilon, y+\epsilon, z+\epsilon)=\epsilon^3$ and varying $\epsilon$ in a ball of radius $r/3$ around $0$. If a proper subset of the coordinates are zero, let $M=\max\{|x|, |y|, |z|\}$ and assume without loss of generality that $y=z=0$ (any other subset of coordinates that are zero yields the same procedure). Then $\phi_3(x, y+\epsilon, z+\epsilon)=x\epsilon^2$ such that varying $\epsilon$ in a ball of radius $r/2$ around 0 yields a ball of radius $M(r/2)^2$ around $B$ in $\phi_3(U)$. If none of the coordinates are zero, then varying $\epsilon$ in a ball of radius $r$ around 0 yields a ball of radius $yzr$ around $B$ in $\phi_3(U)$ since $\phi_3(x+\epsilon, y, z)=B+\epsilon yz$.