When to apply point values in second+ derivative implicit equations

[bernie]

lets say we need to find a 2nd derivative and we have a following equation
$y^4-2x=5$
at point (-2, 1)
so we find the first derivative
$$4y^3\cfrac{dy}{dx}-2=0$$$$4y^3\cfrac{dy}{dx}=2$$
so our first derivative is $\cfrac{1}{2y^3}$

now to the second:
$$\cfrac{-6y^2\cfrac{1}{2y^3}}{4y^6}$$
now here is where I find it a bit confusing. why do we insert the point values (only y in this case, so 1) here and not later?
$$\cfrac{-6*\cfrac{1}{2}}{4} = -\cfrac{3}{4}$$why we don't continue with the following route
$$\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}$$$$\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}=-\cfrac{3}{4y^7}$$fitting 1 in and getting the same $-\cfrac{3}{4}$? Is it always the case and going the first route just saves time?