TITLE: why do I get different derivatives?
QUESTION [6 upvotes]: When I implicitly differentiate with respect to $x$ for the equation $y^{-2}=xy$ I get a different result than when I divide $y$ on both sides first to get $y^{-3}=x$ and then differentiate. 
For the first I get  $dy/dx=-\frac{y^4}{xy^3+2}$ but after dividing both sides by $y$ and then differentiating I get $dy/dx=-\frac{y^4}{3}$ did I break a rule or something or if they are is there a way to show these are equivalent?

REPLY [2 votes]: $y^{-2} = xy\\
-2y^{-3}y' = y + xy'\\
y'(-x-2y^3) = y\\
y' = -\frac {y}{x + 2y^{-3}}$
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
$y' = -\frac {y^2}{3xy}\\
y' = -\frac {y}{3x}$