I am trying to compute a second order Taylor expansion of the following identity: $x = yz$. Where I get stucked is when I transformed the original expression into $y= \frac {x}{z}$, I simply get a different approximation due to the fact that the second order derivative with respect to $z$ in this second form is not zero while they are all equal to 0 in the first form.
What am I missing ? Thanks.
Note that in your setting, in $x=yz$, $y$ is a function in $z$, i.e. $y=y(z)$. So using the Leibnize's rule to calculate the derivatives gives that the fist order derivative of RHS is $$z\frac{dy}{dz}+y(z)$$ and now can you calculate the second order derivative of RHS?