Answer
$$y=\frac{1}{2}x^{\frac{2}{3}}$$
Work Step by Step
$y'=\frac{2y}{3x}$
$\frac{1}{2y}y'=\frac{1}{3x}$
$\int\frac{1}{2y}dy=\int\frac{1}{3x}dx$
$\frac{1}{2}\ln{y}=\frac{1}{3}\ln{x}+C$
$\ln{y}=\frac{2}{3}\ln{x}+C$
$y=e^{\frac{2}{3}\ln{x}+C}=e^Cx^{\frac{2}{3}}$
Let $k=e^C$
so $y=kx^{\frac{2}{3}}$
We need this graph to pass through the point $(8,2)$ so we will apply it as an initial condition.
$2=k8^{\frac{2}{3}}=4k$
so $k=\frac{1}{2}$
and therefore
$$y=\frac{1}{2}x^{\frac{2}{3}}$$
