The problem is not writing compositions properly. If $f$ is a function of $x$ and $y$ and those are functions of $t$ and $u$, then $t \mapsto f(x(t,u),y(t,u))$ is a function of $t$ and $u$, but $f$ itself is not a function of $t$ and $u$, this is an abuse of notation! Rigorously, one defines $\widetilde{f}(t,u)=f(x(t,u),y(t,u))$ and the chain rule says that $$\frac{\partial \widetilde{f}}{\partial t}(t,u) = \frac{\partial f}{\partial x}(x(t,u),y(t,u)) \frac{\partial x}{\partial t}(t,u)+\frac{\partial f}{\partial y}(x(t,u),y(t,u)) \frac{\partial y}{\partial t}(t,u).$$Introducing this extra letter $\widetilde{f}$ just for the sake of being formal is something that people usually don't try to do, so what you're denoting by $\partial f/\partial u$ is actually $\partial \widetilde{f}/\partial u$ for $\widetilde{f}$ defined as above.
If $x$ (for instance) is a function of $t$ and $u$, ${\rm d}x/{\rm d}t$ doesn't make sense, only $\partial x/\partial t$ does. One can make sense of this using the same mechanism above. Namely, define $\widetilde{x}(t) = x(t,u)$ where $u$ is fixed. So $$\frac{{\rm d}\widetilde{x}}{{\rm d}t}(t) = \frac{\partial x}{\partial t}(t,u),$$and $u$ is fixed everywhere throughout this process. One could even denote $\widetilde{x}$ by $\widetilde{x}_u$ (not a partial derivative) to make explicit that a given $u$ has been chosen.
Best Answer
As pointed out by Zkutch, mine is just a misconception. I was falsely considering $\Delta x$ as dependent on $x$. Also, by claiming that $dx$ does not depend on $x$ it was meant that $\dfrac{d^2 x}{dx}=\dfrac{d\Delta x}{dx}=0$ which is true.