Implicit differentiation , leibniz notation- problem differential operator notation in multivariable calc here are three examples of what im finding troublesome in understanding, i can workout the basics by myself but when it comes to iterating the differential operator i find the notation and the process convoluted. My biggest problem is i dont know in which logical step im making the mistake and so i dont know what to search so i can clear my confusion.
example one change the equation $y'' +\frac{2}{x}y'+y=0 $ by taking "$x$ as a function and $t=xy$ for a free variable"
my work- i take $\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\frac{1}{x\frac{\mathrm dx}{\mathrm dt}}$ but my book solution says $\frac{1}{x\frac{\mathrm dx}{\mathrm dt}}-\frac{1}{x^2}$ and i have no idea why since even if they used the chain rule $\frac{\mathrm d (\frac{t}{x})}{\mathrm dt}$ it would not give the book answer.even if we swapped in $y=\frac{t}{x}$ and then using the chain rule i dont get and extra $t$ as in  $\frac{1}{x\frac{\mathrm dx}{\mathrm dt}}-\frac{t}{x^2}$
example two find $\mathrm d z , \mathrm d^2 z$ given implicit function (1) $\frac{x}{z}=\ln(\frac{z}{y})+1$ so i use the total differential on (1) to get $$\frac{z\mathrm dx-x \mathrm dz}{z^2}=\frac{y}{z}\frac{y \mathrm dz - z \mathrm dy}{y^2}$$   from which by rearranging we can get $\mathrm d z$ $$\mathrm d z=\frac{z(y\mathrm dx+z\mathrm dy)}{y(x+z)}$$ $(x\neq-z)$  by simple algebra manipulation, now we use total derivative on  (2)$$ zy\mathrm dx-xy \mathrm dz -yz \mathrm dz-z^2 \mathrm dy=0$$ to get (book answer gives this)$$y(x+z)\mathrm d^2 z=z\mathrm dx \mathrm dy+(z \mathrm dy - x \mathrm dy)\mathrm dz -y^2 \mathrm dz^2$$ and i dont understand this process at all as by going by partial derivatives i dont get the second derivative to look like that(i dont know how i would use chain rule on for example $yz \mathrm dx=z\mathrm dx\mathrm dz+y\mathrm dx\mathrm dz$ ?) and im confused as to what $\mathrm dz^2$ is supposed to mean rigorously at this point. after rearranging and using $\mathrm dz$ the book answer gives $$\mathrm d^2 z=-\frac{z^2(y\mathrm dx-x\mathrm dy)^2}{y^2(x+z)^3}$$
example three  transform $y'''=\frac{6y}{x^3}$ if new variable is $t=\ln(|x|)$ so i begin to calculate $y'$ as follows $$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm dt}\frac{\mathrm dt}{\mathrm dx}=\frac{1}{x}\frac{\mathrm dy}{\mathrm dt}$$ now to calculate the second derivative by t i do as follows $$\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm dy}{\mathrm dt}\right)=\frac{1}{x}\frac{\mathrm dt}{\mathrm dx}\frac{\mathrm d}{\mathrm dt}\left(\frac{\mathrm dy}{\mathrm dt}\right)=\frac{1}{x^2}\frac{\mathrm d}{\mathrm dt}\left(\frac{\mathrm dy}{\mathrm dt}\right)$$ and i thought that was it but the book solution gives this for the $\frac{\mathrm d}{\mathrm dt}(\frac{\mathrm dy}{\mathrm dt})=\frac{\mathrm d^2 y}{\mathrm dt^2}-\frac{\mathrm dy}{\mathrm dt}$ and i dont get how we got to this by using the  chain rule given the first derivative like so $\frac{(\mathrm dy)'\mathrm dt-\mathrm dy (dt)'}{(\mathrm dt)^2}$
im not sure what im getting wrong and i think i need a bit of "hand holding" so i would beg   please explain where im making the mistake or link me to a book/theory for this. i have no formal knowledge in differential forms or wedge products as im a second year math student.
 A: Example one. Since $x$ is a function of $t$ and $y$ is a function of $x$, we have
$$
t=x(t)y(x(t))
$$
or
$$
y(x(t))=\frac{t}{x(t)}.
$$
Applying the chain rule, we get
$$
\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{1\cdot x-t\cdot \frac{dx}{dt}}{x^2}
=\frac1{x}-\frac{t\cdot \frac{dx}{dt}}{x^2}
$$
and
$$
\frac{dy}{dx}=\frac1{x\cdot \frac{dx}{dt}}-\frac{t}{x^2}.
$$
Example two. Technically, the procedure looks like this: the differential from the left and right sides of the equality is taken. The differentials of $dx$ and $dy$ are equal to zero, $d(dz)=d^2z$. We have
$$ 
zy\, dx-xy \, dz -yz \, dz+z^2 \, dy=0.
$$
The first term, $zy\, dx$, gives us $y \,dx\, dz+z\,dx\, dy$; the second, $xy \, dz$, gives $y\,dx\,dz+x\,dy\,dz+xy\,d^2z$, etc.:
$$
y \,dx\, dz+z\,dx\, dy-(y\,dx\,dz+x\,dy\,dz+xy\,d^2z)
$$
$$
-(z\,dy\,dz+y\, dz^2-yz\,d^2z)+2z\,dy\,dz=0
$$
Example three. Here $x$ is a variable, so you shouldn't treat it like a constant:
$$
\frac{d^2y}{dx^2}=\frac{d}{dx} \left( \frac{1}{x}\cdot\frac{dy}{dt}\right)=
-\frac{1}{x^2}\cdot\frac{dy}{dt}+\frac{1}{x}\cdot\frac{d}{dx} \left( \frac{dy}{dt} \right)
$$
