Finding $\partial F/\partial y$ if $F(x,v)=F(u,z)$ and ... I'm here trying to solve the equation (following the solution in Jaynes' Probability Theory book) $F(F(x,y),z)=F(x,F(y,z))$.  Suppose $u = F(x,y)$ and $v=F(y,z)$ and we want $\frac{\partial F(x,v)}{\partial y}$.  We let $F_1(x,y)=\frac{\partial F(x,y)}{\partial x}$ and $F_2(x,y)=\frac{\partial F(x,y)}{\partial y}$.  Here is my calculation, trying to use the chain rule as explicitly as I can in order to find out where I'm going wrong.
$$\frac{\partial F(x,v)}{\partial y} = \frac{\partial F(x,y)}{\partial{x}}\biggr\rvert_{x=x,y=v}\frac{\partial x}{\partial y} + \frac{\partial F(x,y)}{\partial v}\biggr\rvert_{x=x,y=v}\frac{\partial v}{\partial y}$$
Here $\frac{\partial x}{\partial y}=0$ and $\frac{\partial F(x,y)}{\partial v}=1$ and $\frac{\partial v}{\partial y}=F_2(y,z)\frac{\partial y}{\partial y}+\frac{\partial F(x,y)}{\partial z}\frac{\partial z}{\partial y}=F_2(y,z)$ .  Putting this into the equation above,
$$\frac{\partial F(x,v)}{\partial y} = 0+1\cdot F_2(y,z)$$
However, in the book he suggests that the left-hand side is equal to $F_2(x,v)F_1(y,z)$.
Let me know if I'm somehow misusing the chain rule or making any other mistake.
(Note, I think this is pretty neat, it's apparently called The Associativity Equation and was studied by Abel and given an 11-page solution without assuming differentiability by Aczel.  In reasoning from "first principles" about probabilities Jaynes produces it as the necessary form of a probability function.)
 A: Correct application of the chain rule
Differentiating $F(x,F(y,z))$ with respect to $y$ we have:
$$F_2(x,F(y,z))F_1(y,z)$$
That is (applying the chain rule) we differentiate the outer $F$ with respect to its second argument (because there is a $y$ in there), and multiply it by the derivative of the inner $F$ with respect to its first argument ($y$).

Maybe this will help understand how we are applying the chain rule here:
Let $f(y)=F(x,y)$ and let $g(y)=F(y,z)$. Then 


*

*$f(g(y))=F(x,F(y,z))$

*$f'(y)=F_2(x,y)$

*$g'(y)=F_1(y,z)$


We want the derivative of $f(g(y))$. By the chain rule (for compositions of functions of single variables) this is $$f'(g(y))g'(y)=F_2(x,F(y,z))F_1(y,z).$$

Where you went wrong
The first term in your derivative is unnecessary (because as you say it is zero). However, it is not true that what you have called  $$\frac{\partial F(x,y)}{\partial v}\biggr\rvert_{x=x,y=v}$$ is equal to 1. Actually, $$\frac{\partial F(x,y)}{\partial v}\biggr\rvert_{x=x,y=v}=F_2(x,y)$$
