Step in derivation of Euler-Lagrange equations of motion From http://www.mathpages.com/home/kmath523/kmath523.htm

Variations in $x,y,z$ and $X$ at constant $t$ are independent of $t$ (since each of these variables is strictly a function of $t$), so we have
  $${\frac{\partial x}{\partial X}=\frac{\partial{\dot x}}{\partial{\dot X}}}$$

(This is just after equation (5) on the page.)
I'm having trouble making sense of this.  If each of these variables is strictly a function of $t$, and $t$ is held constant, how does a time derivative (${\dot x}$) make sense?
If it were to mean that $t$ is replaced with a constant after the differentiation, then I could take the example of 
$$ x(y)=X^{3/4}=t^{3}$$
$$ X(t)=t^{4}$$
I could then calculate 
$$ {\frac{\partial x}{\partial X}=\frac{\dot x}{\dot X} = \frac{3}{4t}}$$
and 
$$ {\frac{\partial {\dot x}}{\partial {\dot X}}=\frac{\ddot x}{\ddot X}=\frac{1}{2t}}$$
...and those are not equal for any $t$.
What is the justification for this step?
Note: This is a follow up to my first question about this, but that error was resolved.  Rates of change for functions dependent on same variable
 A: *

*Between eq. (4) and eq. (5)


By definition
$$x=x(X(t),Y(t)),$$
and not $x=x(X(t),Y(t),t)$; so
$$\frac{dx}{dt}=\frac{\partial x}{\partial X}\frac{dX}{dt}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},$$
from which follows
$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right)=
\frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt},$$
as $\frac{\partial }{\partial X}\frac{dX}{dt}=\frac{\partial }{\partial X}\frac{dY}{dt}=0.$ On the other hand
$$\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right):=
\frac{d}{dt}g(X(t),Y(t))=
\frac{\partial g}{\partial X}\frac{dX}{dt}+ \frac{\partial g}{\partial Y}\frac{dY}{dt} =  \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt}, $$
i.e.
$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right) =\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right).$$


*

*Just after eq. (5)


As $$x=x(X(t),Y(t)),$$ then $\frac{\partial x }{\partial \dot{X}}=0$ and $\frac{\partial x }{\partial \dot{X}}\left(\frac{\partial x}{\partial X} \right)=0,$ as well. 
Using the notation $\dot{x}=\frac{dx}{dt} $ it follows that
$$\frac{\partial\dot{x} }{\partial \dot{X}}=
\frac{\partial }{\partial \dot{X}}\left(\frac{\partial x}{\partial X}\dot{X}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},\right)=\frac{\partial x}{\partial X},$$
as claimed.
