First of all, the projection $(x,y,z) \mapsto (y,z)$ is an isomorphism between $V$ and $\mathbb A^2$, so you may as well just consider the case $V =\mathbb A^2$, with $(\mathbb C^{\times})^{2}$ acting in the obvious way on the coordinates.
Then the open subset $U = \{ (y,z) \, | \, y z \neq 0\} \subset \mathbb A^2$ (i.e. the complement of the two coordinate lines) is acted on simply transitively by $(\mathbb C^{\times})^2$. Thus $U/(\mathbb C^{\times})^2$ is a point, and since $U$ is open and dense in $V$, to the extent that $V/(\mathbb C^{\times})^2$ has any meaning as a variety, it will be a point too.