Doubts on solving an initial value problem The problem is : $y(\ln y)'= \frac{2t}{y}$ with  $y(0)=0$. How to obtain $y'$ ?

$y(\ln y)'=y(\dfrac{1}{y}\dfrac{dy}{dt})=\dfrac{dy}{dt}.$  
So the problem is $\dfrac{dy}{dt}=\dfrac{2t}{y}\implies ydy=2t dt$ 
So, $\dfrac{y^2}{2}=t^2+C,$ thus with the initial condition $y(0)=0$, we get $C=0$ 
ie, $y^2=2t^2\implies y=\pm \sqrt2t \square$
