I know for instance that $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x)=x^3$ is open. Since $g$ is a homeomorphism, then its inverse $g^{-1}$ is continuous. Let $V \subset \mathbb{R}$ be a open set, then $(g^{-1})^{-1}(V)=g(V) \subset \mathbb{R}$ is open.
But how about $f$? What mechanism can I use to show $f$ is open?
Hint:
$f$ is the composition of the $1$st projection, which is an open map, by the cube function, which is a homeomorphism, hence is an open map.