How to "abstractly" differentiate function, expressed in terms of itself I am interested in differential equations of a single variable in the the dependent function. 
For example 
$$y = e^x \rightarrow \frac{dy}{dx} = y$$ 
$$y = \frac{1}{1-e^{-x}} \rightarrow \frac{dy}{dx} = y(1-y)$$
In general if i'm given $y= f(x)$, How do I find an $F$ such that $y$ satisfies the differential equation
$$ \frac{dy}{dx} = F(y)$$
One solution is 
$$ \frac{dy}{dx} = f'(f^{-1}(y)) $$
Of which the two above are a special case, but how do we characterize the entire set of such  $F$?
 A: If $f$ has a differentiable inverse $g,$ then $x=g(y) .$ And $$dy/dx =(dx/dy)^{-1}=(g'(y))^{-1}= 1/ (f^{-1})' (y).$$
A: This is a specific form of a separable ODE. In general, if you have an equation of the form $\frac{dy}{dx} = f(x)g(y)$, you rearrange it into the form $\frac{1}{g(y)}\frac{dy}{dx} = f(x)$ and integrate with respect to $x$, giving:
$\int{\frac{1}{g(y)}\frac{dy}{dx}\ dx} = \int{\frac{1}{g(y)}\ dy}= \int{f(x)\ dx}$
Where the transition in the first equality is essentially just applying integration by substitution/change of variables.
A: If $\frac{dy}{dx}= F(y)$ then $\frac{dy}{F(y)}= dx$ so, as long as you can integrate F(y) you can solve for x as a function of y and then try to invert the function: $\int dx= x= \int\frac{dy}{F(y)}$.
In the examples you give, we can write $\frac{dy}{dx}= y$ as $\frac{dy}{y}= dx$ so $\int \frac{dy}{y}= ln|y|= x+ C$ and, taking the exponential of both side $y= C'e^x$. (Strictly speaking we would have "$|y|= e^Ce^x$" but since we are free to take C' positive or negative we get y itself.) 
Similarly, if $\frac{dy}{dx}= y(1- y)$ then $\frac{dy}{y(1- y)}= dx$.  We can use "partial fractions" to write $\frac{1}{y(1- y)}$ as $\frac{1}{y}- \frac{1}{y- 1}$.  Integrating, $\int \left(\frac{1}{y}- \frac{1}{y- 1}\right)dx= ln(y)- ln(y-1)= ln\left(\frac{y}{y- 1}\right)= \int dx= x+ C$.  Taking the exponential of both side again, $\frac{y}{y- 1}= C'e^x$.  Solving that for y, $y= \frac{C'e^x}{C'e^x- 1}$.  Multiplying both numerator and denominator by $e^{-x}$, $y= \frac{C'}{C- e^{-x}}$ and you get your "$y= \frac{1}{1- e^{-x}}$" by taking C'= 1.
