why do I get different derivatives?

These are the same.

Note that $$x=y^{-3}\implies xy^3+2=3$$

$y^{-2} = xy\\ -2y^{-3}y' = y + xy'\\ y'(-x-2y^3) = y\\ y' = -\frac {y}{x + 2y^{-3}}$

Rather than simplifying this to $y' = -\frac {y^4}{xy^3 + 2y}$

lets say

$y' = -\frac {y^2}{xy + 2y^{-2}}$

Now we can substitute from the original equation.

$y' = -\frac {y^2}{3y^{-2}}\\ y' = -\frac {y^4}{3}$

or

$y' = -\frac {y^2}{3xy}\\ y' = -\frac {y}{3x}$