Find $d^2z$ of the following function $$\frac{x}{z}= \ln{\frac{z}{y}}+1$$
Finding the total derivative of the left and right side we arrive at the expression:
$$\frac{zdx-xdz}{z^2}=\frac{y}{z} \cdot \frac{ydz-zdy}{y^2} \iff yzdx-xydz-yzdz+z^2dy=0$$
Differentiating again this expression:
$$y(x+z) d^2z=zdzdy+(zdy-xdy)dz - y^2dz^2$$
Also $dz=\frac{z(ydx+zdy)}{y(x+z)}$ so by substituting in the last equation we get:
$$d^2z= - \frac{z^2(ydx-xdy)^2}{y^2(x+z)^3}$$
I have couple of questions regarding this solution. I've managed to replicate taking the first derivative from both sides of the equation and arrived at the same result. What I noticed, is that during this we do not consider $z$ as a function of $x$ and $y$.
Next when taking the second derivative I didn't get the extra term $y(x+z)d^2z = yx d^2z +z d^2z$. This term seems to appear by differentiating all the $dz$ parts in $yzdx-xydz-yzdz+z^2dy=0$. Perhaps it is from somewhere else that we get the $d^2z$ term but I don't know what I'm missing.