enter link description here You may see through the link that someone pointed out that if$$f=f(x(t),y(t))$$then$$\cfrac{\partial f}{\partial t}=0$$So now i want to challenge this by giving an example:
Let $f=f(x,y)$, where $x=x(t)$,$y=y(s)$. Because $f$ is independent of $t$, so we have $$\frac{\partial f}{\partial t}=0\qquad(1)$$
Now, given that $x=x(t)=2t^2$, $y=y(s)=3s$, then f is given by:$$f=f(x,y)=xy\qquad(2)$$ Substitute $x=x(t)=2t^2$, $y=y(s)=3s$ into (2), we have$$f=6t^2s\qquad(3)$$It follows that $$\frac{\partial f}{\partial t}=12ts\qquad(4)$$Equation(1) and Equation (4) are not equal, what is wrong?
You are not understanding what people said in answering your other question.
The root of your confusion is that you write things like $f=f(x(t), y(t)$ without making clear what this means.
The true story is that there is a function $f:\mathbb{R}^2\to\mathbb{R}$, say $(x,y)\mapsto f(x,y)$.
Then there are two functions $x:\mathbb{R}\to\mathbb{R}$, say $t\mapsto x(t)$: and $y:\mathbb{R}\to\mathbb{R}$, say $t\mapsto y(t)$.
From these we can form a new function, $F:\mathbb{R}\to\mathbb{R}$, given by $t\mapsto F(t)=f(x(t), y(t))$. You can then sensibly partially differentiate $f$ with respect to $x$ and $y$; differentiate each of $x$ and $y$ with respect to $t$; and differentiate $F$ with respect to $t$.
The Chain Rule then says $$ \frac{dF}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt} +\frac{\partial f}{\partial y}\frac{dy}{dt}.$$
Warning
It leads to endless confusion to use the same letter for both $f$ and $F$. I know that people do this, but they are confusing values with functions.
In the "example" you give you confuse things further by jumping from $f=f(x(t),y(t))$ to $f=f(x(s),y(t))$ as though this made any sense. Here you've got yet another function, $\phi:\mathbb{R}^2\to\mathbb{R}$ given by $\phi(s,t)=f(x(s),y(t))$. Using the Chain Rule correctly will remove your confusions/supposed contradictions.