If $f=f(x(s),y(t))$, then $\frac{\partial f}{\partial t}=0$ is wrong? 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?
 A: 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.
A: $f$ is not independent of $t$.
$f$ depends on $x$ and $y$, both of which in turn depend on $t$. So $f$ depends on $t$.
Remember the chain rule:
$$f_t = f_xx_t +f_yy_t$$
A: $t$ affects $f$. in other words $f$ and $t$ is dependent since $f$=$f$($x$($s$),$y$($t$)) hope this helps! :)
