Geometric Intuition for ∂y'/∂y = 0 According to this question, in the sense that the geometry of $y(x)$ is a curve in the plane, the geometry of $z(x,\ y(x))$ is a curve in $3$-space.  Suppose we consider such a function $z$ whose value happens to be given by $\frac{dy}{dx}$.  In order to agree with Wolfram's proclamation that $\frac{\partial \frac{dy}{dx}}{\partial y} = 0$, it should be the case that at every point along the curve, a tiny nudge parallel to the $y$-axis results in no change in the height of the curve.
Now, either the curve is oriented such that our tiny nudge keeps us on the curve, or it isn't and we instead land in the surrounding ambient space.  In the former case, is there a way to see why we could never increase or decrease in height in response to such a nudge, in order to illustrate why $\frac{\partial \frac{dy}{dx}}{\partial y} = 0$?  In the latter case, does taking a partial derivative with respect to $y$ even make sense, and if so, how should it be thought of geometrically?
 A: For arbitrary $z(x,y)$, we can clearly have $\dfrac{\partial}{\partial y}z(x,y) \neq 0$. Because, why not?
However, if you constrain $y$ to $f(x)$, then under that constraint $y$ is not allowed to vary unless $x$ does, because $y$ is now given as an explicit function of $x$. To express this constraint we may write $z(x,y)\big{|}_{y=f(x)}$ or more simply $z\big{(}x,f(x)\big{)}$.
A partial-derivative with respect to not-$x$ requires holding $x$ fixed. This means that $\dfrac{\partial}{\partial y}z\big{(}x,f(x)\big{)} = 0$, because again, the constraint forces that fixed $x$ means constant $y$. Basically, the $\dfrac{\partial}{\partial y}$ tried to wiggle $y$, but the constraint said "$y$ is not relevant here, there is only $x$" and so $\dfrac{\partial}{\partial y}$ reported $0$.
To visualize an example, here is a plot of $z(x,y)=y$, which has $\dfrac{\partial}{\partial y}z(x,y) = 1$,

and here is a plot of that $z$ constrained, $z(x,y)\big{|}_{y=x^2}=x^2$, which must have $\dfrac{\partial}{\partial y}z\big{(}x,x^2\big{)} = 0$,

The curve you have conceptualized is the intersection of these two surfaces.


... a tiny nudge parallel to the -axis results in no change in the height of the curve.

I don't think that makes sense because "moving" parallel to the $y$-axis without changing $x$ doesn't necessarily keep you "on" the curve (i.e. it can violate the constraint). The thing that is really not changing height when you move in the $y$-direction is the surface $g(x,y)=z\big{(}x,f(x)\big{)}$, like in the second plot above.
I think your confusion resulted from the notation "$z\big{(}x,y(x)\big{)}$" as opposed to "$z\big{(}x,f(x)\big{)}$". There is a conceptual difference between $y$ the dimension and $y$ the 2nd-argument-of-the-function $z:\mathbb{R}^2 \to \mathbb{R}$.
Unrelated to all this constrained $z(x,y)$ business, "Wolfram's proclamation" is indeed true in general. We can see that the definition of $\dfrac{dy}{dx}$ is invariant to variations of $y$,
\begin{align*}
\dfrac{dy}{dx} &:= \lim_{\delta_x \to 0} \dfrac{y(x+\delta_x) - y(x)}{\delta_x} \tag{$x, y, \delta_x \in \mathbb{R}$}\\\\
&= \lim_{\delta_x \to 0} \dfrac{\big{(}y(x+\delta_x) + \delta_y\big{)} - \big{(}y(x) + \delta_y\big{)}}{\delta_x} \tag{$\delta_y \in \mathbb{R}$}\\\\
&= \dfrac{d(y+\delta_y)}{dx}
\end{align*}
It then follows that,
\begin{align*}
\dfrac{\partial}{\partial y}\dfrac{dy}{dx} &= \lim_{\delta_y \to 0} \dfrac{\dfrac{d(y+\delta_y)}{dx} - \dfrac{dy}{dx}}{\delta_y}\\\\
&= \lim_{\delta_y \to 0} \dfrac{\dfrac{dy}{dx} - \dfrac{dy}{dx}}{\delta_y}\\\\
&= 0 \tag{L'Hopital}
\end{align*}
