Having no luck with this either
Find the solution of the initial value problem
dy/dx=(1/2e^-x/4).(y sqrt y) (y>0) y=1/4 when x =0
do you mean $\displaystyle y' = \frac 12e^{-x/4} \cdot y \sqrt{y}$ ?
if so, this si separable, divide both sides by $\displaystyle y \sqrt{y} = y^{3/2}$ (we can do this since $\displaystyle y \ne 0$). then, multiply both sides by $\displaystyle dx$, we get
$\displaystyle \Rightarrow y^{-3/2}~dy = \frac 12e^{-x/4}~dx$
now integrate both sides and continue