Question re: defining exponential function On page 212 of "Burn Math Class," we're presented with the following line of reasoning (note that $x$ and $y$ are simply two numbers and neither is a function of the other):
1) Define a function by the behavior $f(x + y) = f(x) \cdot f(y)$ 
2) We can deduce that $f(0) = 1 $
3) We differentiate with respect to y and get $f'(x + y) = f'(y) \cdot f(x)$ 
4) By setting $y=0$ in line #3 we find: $f'(x) = f'(0) \cdot f(x)$
Here are my questions (I sent these to the author directly, but got no reply):

A) In differentiating the equation (step #3 above) the author mentions we need to make use of two facts: $\frac{dx}{dy} = 0$ and $\frac{d(x + y)}{dy} = 1$. I can't see where the second expression gets used, as I simply apply the product rule to get the derivative. What am I missing?
B) Because $\frac{dx}{dy} = 0$, it seems like the derivative in line #4 would always equal $0$. Here's why: 

i) $\frac d{dy} [f(x+y)] = f(x) \frac d{dy} [f(y)]$ 
ii) With $y=0$ this simplifies to $\frac d{dy} [f(x)] = f(x) \cdot \frac d{dy} [f(0)]$ 
iii) By the chain rule, $\frac d{dy} [f(x)] = \frac {df(x)}{dy} \cdot \frac{dx}{dx} = \frac{df(x)}{dx} \cdot \frac{dx}{dy} = \frac {df(x)}{dx} \cdot 0 = 0$. Clearly something is wrong here, but I can't see what.

C) We conclude the section by noting that the expression $f'(x) = f'(0) * f(x)$ means this function's derivative is almost equal to its first derivative. But we seem to have glossed over the fact that we differentiated with respect to $y$, not $x$ (such is the drawback of prime notation). In other words, what we really have is $\frac d{dy} [f(x)] = f(x) *\cdot \frac d{dy} [f(0)]$. Maybe this doesn't matter because $x$ and $y$ are just arbitrary numbers, but something doesn't quite sit right with me.

I apologize for the long post - hopefully someone has the patience and insight to help answer my questions.
Edit: here's a link to the page in question - reading the original text might help clarify my questions. https://ibb.co/kePfiv
 A: A) $\frac{df(x+y)}{dy} = \frac{df(x)f(y)}{dy}$
$= \frac{df(x)}{dy}f(y) + f(x)f'(y)$ but we don't know what $ \frac{df(x)}{dy}$ is unless we use the chain rule.  $\frac{df(x)}{dy} = f'(x)*\frac{dx}{dy} = f'(x)*0 = 0$. so $\frac{df(x+y)}{dy}= f(x)f'(y)$.  But we don't know that $\frac{df(x+y)}{dy}= f'(x+y)$.
Using the chain rule was get $\frac{df(x+y)}{dy} = f'(x+y)\frac{d(x+y)}{dy} = f'(x+y)*1= f'(x+y)$.
B) You can't take a derivative in terms of a variable when the variable is fixed.  That's simply a meaningless concept.
C) A derivative is a derivative.  If $f'(z) = g(z)$ then $f'(pinkhonkhonk) = g(pinkhonkhonk)$ it doesn't matter what variable label we used to derive it.  We did some fancy footwork to discover that if $f$ has the property that $f(a+b) = f(a)f(b)$ then $f$ also has the very weird property that $f'(a+b)=f'(b+a)= f(a)f'(b)=f(b)f'(a)$.  We found out that result but differentiating $f(x+y)$ in terms of $y$ but the result is one about $f$ and $f'$ for any values.
You state "what we really have is $\frac d{dy} [f(x)] = f(x) *\cdot \frac d{dy} [f(0)]$".  Not quite.  What we really have is $\frac d{d(x+y)} [f(x+y)]_{x+y = x}*\frac{d{x+y}}{dy}_{y=0} = f(x) *\cdot \frac d{dy} [f(y)]_{y=0}$
And as
$\frac d{d(x+y)} [f(x+y)]_{x+y = x} = f'(x+y)_{x+y=x} = f'(x)$.
$\frac{d{x+y}}{dx}_{y=0} = 1_{y=0} = 1$
$\frac d{dy} [f(y)]_{y=0}=f'(y)_{y=0} = f'(0)$.
That is nothing more or less than $f'(x) = f'(0)f(x)$
