Answer
$$y=\frac{1}{3}\sqrt{x}$$
Work Step by Step
$y'=\frac{y}{2x}$
$\frac{y'}{y}=\frac{1}{2x}$
$\int\frac{1}{y}dy=\int\frac{1}{2x}dx$
$\ln{y}=\frac{1}{2}\ln{x}+C$
$e^{\ln{y}}=y=e^{\frac{1}{2}\ln{x}+C}=e^Ce^{\frac{1}{2}\ln{x}}$
Let $k=e^C$
then $y=k\sqrt{x}$
Now we need our graph to pass through the point $(9,1)$ so we will apply it as an initial condition.
$1=k\sqrt{9}$
$\frac{1}{3}=k$
and so $$y=\frac{1}{3}\sqrt{x}$$
