Relationship between $t$ and $x$ in $x'=f(t,x)$? I'm a couple weeks into a course on differential equations. I've been told that it's more accurate to think of a differential equation $x'=f(t,x)$ as being determined by two separate variables, $t$ and $x$. So, my understanding is that while it is true that for a particular solution of the DE, say $x(t)$, the dependence of $x$ on $t$ is evident, it is more accurate to view $x'=f(t,x)$ as a statement about two separate variables.
Okay! So I'm holding $t$ and $x$ as distinct in my mind, and I go to solve a basic differential equation:
$$x'=-2tx^2$$
$$-\frac{x'}{x^2}=2t$$
And it's at this point that I suddenly remember a fortuitous fact: There is an implicit dependence of $x$ on $t$, after all! This allows me to integrate with respect to $t$, appealing to the chain rule and the FTC I: If $F(x)$ is an antiderivative of $-\frac{1}{x^2},$ then $\frac{d}{dt}(F(x(t)))=-\frac{x'}{x^2}$ by the chain rule, and $\int{-\frac{x'}{x^2}dt}=\int{\frac{d}{dt}F(x(t))}=F(x(t))$ by FTC I (since $\frac{1}{x^2}$ is continuous everywhere it is defined). So,
$$\int{-\frac{x'}{x^2}dt}=\int{2t dt}$$
$$\frac{1}{x}=t^2+C$$
$$x(t)=\frac{1}{t^2+C}.$$
It's only the very real dependence of $x$ on $t$ that allows me to use the chain rule and the FTC I to justify all that integrating. So, is the $x$ in $x'=f(t,x)$ truly a second variable? Another question is, Does it even matter? Like anyone, I can sometimes make mountains of molehills when I'm learning something new, and create difficulties for myself where none exists (as if things weren't difficult enough!). Is that what I'm doing here? Does it truly matter how I think of the relationship between $t$ and $x$ in the statement $x'=f(t,x)$?
 A: 
I've been told that it's more accurate to think of a differential
  equation $x'=f(t,x)$ as being determined by two separate variables,
  $t$ and $x$.

This may be because a lot of theorems require you to see if $f$ itself satisfy some hypotheses to guarantee uniqueness/existence etc. Now, $f$ determines a differential equation, but it does not care about the differential equation by itself, and that is what you have been implicitly told. Let's work out an example so as to make more precise what you are saying (or what you've been told) means.
Consider your example:
$$x'=-2tx^2.$$
I can consider the function $f: \mathbb{R}^2 \to \mathbb{R}$ given by $f(a,b)=-2ab^2$. This function does not care about differential equations at all. However, when I compose it with the "lift" (for lack of a better name. Call it whatever you like.) of a differentiable function $x$, i.e., the map $\widetilde{x}:t \mapsto (t,x(t))$, we get your differential equation.
Explicitly, the differential equation is:
$$x'=f \circ \widetilde{x}.$$
The distinction I am trying to make is: when you are actually trying to solve a differential equation, the differential equation obviously couples $t$ and $x$. However, $f$ doesn't. We have that $f$ determines a differential equation, but is not a differential equation by itself. It is blind to it, it is a just a function of two variables. As any function of two variables, one is "independent" of each other.
