Is the property "being a derivative" preserved under multiplication and composition? Since differentiation is linear, we therefore have that if $f, g: I\to \mathbb{R}$ is a derivative (where $I\subset \mathbb{R}$ is an interval), then so does their linear combination. What if we consider their multiplication and composition?
Due to the forms of the product rule of differentiation of product function and chain rule of differentiation of composition, I highly doubt their product or composition necessarily is still a derivative, but I cannot construct counterexamples.

Let me address just one of your problems.

Problem. Suppose that $f$ and $g$ are both derivatives.   Under 
   what conditions  can we assert that  the product $fg$ is
  also a derivative?

The short answer is that this is not true in general.  In fact even if we assume that $f$ is continuous and $g$ is a derivative the product need not
be a derivative.  However if we strengthen that to assuming that $f$ is not merely continuous but also of bounded variation, then indeed the product with any derivative would be a derivative.
This is an interesting problem and leads to interesting ideas.
For references to the statements here and an in depth look at the problem here are some references:

Bruckner, A. M.; Mařík, J.; Weil, C. E. Some aspects of products of
   derivatives. Amer. Math. Monthly 99 (1992), no. 2, 134–145.
Fleissner, Richard J. Distant bounded variation and products of
    derivatives . Fund. Math. 94 (1977), no. 1, 1–11.
Fleissner, Richard J. On the product of derivatives. Fund. Math. 88
  (1975), no. 2, 173–178.
Fleissner, Richard J. Multiplication and the fundamental theorem of
  calculus—a survey. Real Anal. Exchange 2 (1976/77), no. 1, 7–34.
Foran, James 
  On the product of derivatives, Fund. Math. 80 (1973), no.
  3, 293–294.

I will edit in some links when I find them.  Foran and Fleissner were close childhood friends who ended up pursuing their PhD at the same time in Milwaukee.  Fleissner died in an automobile accident in 1983.
NOTE ADDED.  Elementary students are not going to want to pursue this topic to quite this depth.  But here is an exercise aimed at this level that they might find entertaining.

Exercise.  Consider the function $$f(x)=\begin{cases} \cos \frac1x, & x\not=0 \\ 0 &x=0 \end{cases} $$  Show that the function $f$ is a
  derivative but that its square $f^2$ is not.

