The given equation of the parabola is $(x-h)^2=4py$. Since it has two constants we will have to differentiate twice.
Differentiating wrt x we get
$2(x-h)=4py'$
I.e. $(x-h)=2py' $
Differentiating once again wrt x we get
$1=2py'' $
Therefore $y''=\large\frac{1}{2p}$
Hence $(x-h)=\large\frac{y'}{y''}$
Substituting this in the first equation we get $ (\large\frac{y'}{y''})^2= (x-h)^2 = 4py = 2p*2y = \large\frac{2y}{y''} $
Therefore on simplifying we get
$(y')^2-2y(y'')=0 $