Consider a function $f(x,y,z(x,v))$. Does $\frac{\partial f}{\partial z}$ hold $x$ and $v$ constant? where $x,y,v$ are independent variables.
Basically, I am confused with the fact that a partial derivatives holds all variables constant, other than the variable with which we are taking the derivative, but here we are taking the derivative with respect to a function of other variables.
In other words, what is being held constant in $\frac{\partial f}{\partial z}$. Perhaps this partial derivative is telling us what the change would be if we could change $z$ without changing $x$ and $v$?
 A: Yeah, $\frac{\partial f}{\partial z}$ doesn't take into account the fact you are pre-composing $f$ with another function. It really should be written $D_3 f$, since it doesn't matter which letter you used when you defined what the function would to with its third entry. All that matters is that you are differentiating the third entry while keeping the other two constant.
That said, the assignment $(x,y,v)\mapsto f(x,y,z(x,v))$ is NOT the function $f$. It is a new function (call it $h$ if you want) which you construct by precomposing the third entry of $f$ with another function called $z$ (note that, if you want to be consistent, once have a function called $z$ you can no longer use the same symbol for a variable, e.g. as in $\frac{\partial}{\partial z}$). This new function $h$ has three entries again, $h(x,y,v)=f(x,y,z(x,v))$, and you can differentiate with respect to its first, second or third entry.
A: The function $f$ doesn't "know" that its third argument depends on other variables - all it knows is that it takes three arguments and gives a number back. $\frac{\partial f}{\partial z}$ is a "question" you're asking $f$: "if your third argument were to change without the other two changing, how would your output change?" That question is about what could go into $f$, not about what is actually going into $f$.
