I'm given the equation: $xz^2+\frac{y}{z}-2xyz=3$ and I'm asked to find all the second derivatives of $z$ as a function of $x$ and $y$ at the point $(1,2,-1)$.
first I define a function $f(x,y,z) = xz^2+\frac{y}{z}-2xyz$ and in order to find the first derivative I'll use the differential of $f$. Therefore:
$$z_x'= \frac{\partial z}{\partial x}= -\frac{f_x'}{f_z'} = -\frac{z^2-2yz}{2xz-\frac{y}{z^2}-2xy}$$ and so on with $z_y'= \frac{\partial z}{\partial y}$
When I look for a second derivative I actually use the chain rule of derivatives, so basically: $$z_{xx}''= \frac{\partial }{\partial x}\bigl(-\frac{z^2-2yz}{2xz-\frac{y}{z^2}-2xy}\bigr)$$ Now since I defined (or was asked to consider) $z$ as $z(x,y)$ every time I derive $z$ with respect to $x$ I have to multiply it by $z_x'$ I found earlier.
My question is: since implicit functions let me represent each variable as a function of all the others, won't I have to do the exactly same thing with $y$ with respect to $x$ and vice versa?
For example: $$z_{xx}''=^? -\frac{(2z-2y)z_x'\times(2xz-\frac{y}{z^2}-2xy)-(2z-\frac{2y}{z^3})z_x'\times(z^2-2yz)}{(2xz-\frac{y}{z^2}-2xy)^2}$$
Now when I derive with respect to $x$, I also derive things with $y$ in them, should I according to the chain rule, have to multiply it by $y_x'$? According to the logic of implicit function where I can consider of each variable as a function of the rest it feels like a possibility to consider, although I'm asked to focus on $z$ as function of $x$ and $y$ so I can use that chain rule so confidently.
Any insight on the matter would be much appreciated.