Answer
$$\eqalign{
& \,{\text{Vertex }}\left( {0,0} \right) \cr
& \,\,\,\,\,\,\,{\text{Focus: }}\left( {0,p} \right):\,\,\left( {0,\frac{1}{{12}}} \right) \cr
& \,\,\,\,\,\,\,{\text{Directrix:}}\,\,y = - p:\,\,\,\,\,y = - \frac{1}{{12}} \cr
& \,\,\,\,\,\,{\text{Vertical axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\,\,\,\,\left[ {0,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& 3{x^2} = y \cr
& {x^2} = \frac{1}{3}y \cr
& {\text{The equation is written in the form }}{x^2} = 4py \cr
& {x^2} = \frac{1}{3}y \to \,\,\,\,\,\,\frac{1}{3} = 4p\,\,\,\,\,\,\,\,p = \frac{1}{{12}} \cr
& {\text{With: }} \cr
& \,\,\,\,\,\,\,{\text{Vertex }}\left( {0,0} \right) \cr
& \,\,\,\,\,\,\,{\text{Focus: }}\left( {0,p} \right):\,\,\left( {0,\frac{1}{{12}}} \right) \cr
& \,\,\,\,\,\,\,{\text{Directrix:}}\,\,y = - p:\,\,\,\,\,y = - \frac{1}{{12}} \cr
& \,\,\,\,\,\,{\text{Vertical axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\,\,\,\,\left[ {0,\infty } \right) \cr} $$
