Answer
$\mathrm{Decreasing:}\:\:\:(-\infty,0)$
$\mathrm{Increasing:}\:\:\:(0,\infty)$
Graph is symmetric with respect to the $\ \mathrm{y-axis}.$
Work Step by Step
$\mathrm{First\:Part:}\:\:$ According to the definitions:
A function $\ f\ $ defined on an interval is increasing on $\ (a, b)\ $ if for every $\ x_1, x_2\ $ $\in$ $(a, b)$ $\ x_1\le x_2\ $ implies that $\ f(x_1)\le f(x_2).\ $
A function $\ f\ $ defined on an interval is decreasing on $\ (a, b)\ $ if for every $\ x_1, x_2\ $ $\in$ $(a, b)$ $\ x_1\le x_2\ $ implies that $\ f(x_1)\ge f(x_2).\ $
We can write the given function $\ y=(-x)^{\frac{2}{3}}\ $ as $\ y=\sqrt[3] {(-x)^2}=\sqrt[3] {x^2}.$
First of all, create a table with a few points to sketch the graph.
$\quad \mathrm{See\:the\:table\:and\:graph\:above.}$
$\mathrm{Second\:Part:}\:\:$ Graph is symmetric with respect to the $\ \mathrm{y-axis}.$
$\mathrm{Third\:Part:}\:\:$ The graph of the given function $\ y=\sqrt[3] {x^2}\ $ is decreasing on $\ (-\infty,0)\ $ and is increasing on $\ (0,\infty).$
