Dependence of variables - partial differentiation When doing partial differentiation, I'm having trouble seeing if an expression depends on a variable or not. For example, consider the following expressions:
$$a = x + 2y, $$
$$b = 3x -y,$$
$$z = 3a + 4.$$
So, by blindly differentiating, I can say
$$\frac{\partial z}{\partial b} = 0,$$
since $z$ has no (explicit!) dependence on $b$. However, intuitively, if $b$ changes, $x$ and $y$ may be changing, leading to a change in $a$, and as a result, a  change in $z$, implying that the partial derivative may not be $0$. 
How do I reconcile these two notions?
 A: Disclaimer: I find this question relatable, so I am going to answer what works for me, although there may be some hand-waving...
The problem has to specify the cause-effect relation among variables. I usually do this with a sequence of transformations represented by arrows. For instance, you can say that you have a couple of fundamental variables $(x,y)$, which are transformed to other pair $(a,b)$ (in this case it is a linear change of basis), which in turn can be used to produce the magnitude $z$. So you have the chain:
$$z\gets (a,b) \gets (x,y)$$
If we assume that this is what is meant by your three equations, then it is true that a change in $(x,y)$ forces a change in $b$, but $z$ is only aware of this because of the change in $a$, so $\frac{\partial z}{\partial b}=0$.
However, imagine we start from the assumption that the fundamental variables are $(a,b)$ and there is the chain:
$$z \gets (x,y) \gets (a,b) $$
where the function $(x,y)=F(a,b)$ is implicit. 
Then, your third equation is a (notationally abusive) way to express:
$$z=3(x+2y)+4$$
so you can apply the chain rule to obtain:
$$
\frac{\partial z}{\partial b}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial b}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial b}
$$
To summarize: the relations among variables must be stated as a part of the problem.
A: You need to point out what your variables are. If your variables are $a$ and $b$, and  you regard $z$ as $z(a,b)=3a+4$, then $z(a,b)$ is of course independent on $a$.
If you choose $x,y$ as variables, you should write $z=3a(x,y)+4$ to tell yourself that $a$ is the function of variables $x$ and $y$.
A: $\frac{\partial z}{\partial b} = 0$ is only true if $\frac{\partial a}{\partial b} = 0$ which it is not given the equations that govern $a$ and $b$. Thus, you made a mistake when you concluded that $\frac{\partial z}{\partial b} = 0$.
