Does the chain rule imply the product rule? Let $\mathbb{F}$ be a field, and consider $\mathbb{F}^\mathbb{F}$ as an algebra over $\mathbb{F}$ with the standard function multiplication. Let $D$ be a linear transformation on a subalgebra of $\mathbb{F}^\mathbb{F}$ closed under function composition that satisfies the chain rule.  Does $D$ necessarily satisfy the product rule for arbitrary $\mathbb{F}$? (Inspired by a comment on this question.) What if the subalgebra must be unital?
 A: Let $\mathbb{F}=\mathbb{F}_2$ and consider $D:\mathbb{F}_2^{\mathbb{F}_2}\to\mathbb{F}_2^{\mathbb{F}_2}$ which sends the constant functions to $0$ and the nonconstant functions to $1$.   It is easy to check that this satisfies the chain rule but it does not satisfy the product rule since $D(i\cdot i)=D(i)=1\neq 2iD(i)=0$ where $i$ denotes the identity function.
On the other hand, if $\mathbb{F}$ has characteristic different from $2$, then the chain rule actually does imply the product rule (assuming your subalgebra $A\subseteq\mathbb{F}^\mathbb{F}$ is unital and contains the identity function $i$).  First, note that $$D(1)=D(1\circ 0)=(D(1)\circ 0)\cdot D(0)=0$$ since $D(0)=0$.  Also, for any $f\in A$, $$D(f)=D(f\circ i)=D(f)D(i).$$ Taking $f=i$ gives $D(i)=D(i)^2$, so $D(i)$ can only take the values $0$ and $1$.  Let $S\subseteq\mathbb{F}$ be the set of inputs on which $D(i)$ is $0$; then $D(f)$ vanishes on $S$ for all $f\in A$.  For each $a\in\mathbb{F}$, let $s_a$ be the function $a-i$.  Note that $s_a\circ s_a=i$ so $$D(i)=(D(s_a)\circ s_a)\cdot D(s_a).$$  Comparing the vanishing sets of each side of this equation, we see that $S\supseteq s_a^{-1}(S)=s_a(S)$.  Since $a\in\mathbb{F}$ is arbitrary, this means $S$ must be either $\emptyset$ or all of $\mathbb{F}$.  If $S=\mathbb{F}$ then $D(i)=0$ and thus $D=0$ and the conclusion is trivial.  So let us assume $S=\emptyset$, which means $D(i)=1$.
Now let $f=D(i^2)$.  For any $a,b\in\mathbb{F}$, we have $$D((ai+b)^2)=(f\circ (ai+b))D(ai+b)=af\circ(ai+b)$$ but also $$D((ai+b)^2)=D(a^2i^2+2abi+b^2)=a^2f+2ab.$$  That is, for each $x\in\mathbb{F}$, $$af(ax+b)=a^2f(x)+2ab,$$ or $$f(ax+b)=af(x)+2b$$ as long as $a\neq 0$.  Plugging in $x=0$ gives $$f(b)=af(0)+2b.$$  Since $\mathbb{F}$ has more than $2$ elements (so there are multiple different choices for $a$) this implies $f(0)=0$ and thus $f(b)=2b$.  That is, $f=2i$.
Now let $f,g,\in A$ be arbitrary and consider $D((f+g)^2)$.  On one hand, $$D((f+g)^2)=(2i\circ (f+g))\cdot D(f+g)=2(f+g)D(f+g).$$  On the other hand, $$D((f+g)^2)=D(f^2+2fg+g^2)=2fD(f)+2D(fg)+2gD(g).$$  Since $2\neq 0$, comparing these two equations gives $D(fg)=fD(g)+gD(f)$, as desired.
(I don't know what can be said if $\mathbb{F}$ has characteristic $2$ but more than $2$ elements.)
A: Not an answer, but a helpful heuristic.
The chain rule and the product rule cannot be compared. A product of functions, which you called standard, requires two functions with the same domain and range: $f,g\, : \,D\rightarrow R$ since we plug in the same value and multiply in a set where multiplication has to be defined:
$$(f\cdot g)(x)=f(x)\cdot g(x).$$
On the other hand, a composition of functions requires that the range from one function is within the domain of another: $f\, : \,R\rightarrow S$ and $g\, : \,D \rightarrow R$ in order to make
$$
(f\circ g)(x)=f(g(x))
$$
possible. If you simplify all these requirements by setting e.g. $D=R=S=\mathbb{R}$ then you disguise those subtleties. They are still there. Hence product and composition are two very different operations. It is even more obvious if we consider the derivatives:
\begin{align*}
D_p(f\circ g)&=D_{g(p)}(f)\cdot D_p(g)\\
D_p(f\cdot g)&=D_p(f)\cdot g+ f\cdot D_p(g)
\end{align*}
The fact that the evaluation points differ significantly makes them incomparable.
