Partial derivatives in a function with "linked" variables Let $x,y,z$ be three variables, $\phi=\phi(x,y,z)$ be a function of these variables and $c_1,c_2$ be two constants. If we are given that
$$x=c_1\left(\frac{y}{z}\right)^{1/3} \tag{1}$$
$$y=c_1^{-3}x^3z \tag{1'}$$
$$\phi(x,y,z)=c_2zx^4 \tag{2}$$
and we want to find $\frac{\partial\phi(x,y,z)}{\partial z}$, then we get that
$$\frac{\partial\phi(x,y,z)}{\partial z}=\frac{\partial (c_2zx^4)}{\partial z}=c_2x^4+c_2z\frac{\partial (x^4)}{\partial z} \tag{3}$$
But, to find $\frac{\partial (x^4)}{\partial z}$, we would need to use $(1)$ to write the dependence of $x$ on $z$. However, by doing that, we would need to find $\frac{\partial (y^{3/4})}{\partial z}$, which would depend on $z$ by $(1')$, so we would have the same problem ad infinitum.
How are partial derivatives calculated in case the variables are linked, as in this example?
 A: It should just be $$\frac{\partial\phi(x,y,z)}{\partial z} = c_2x^4.$$  We can see this from the definition of partial derivative:
\begin{align}
\frac{\partial\phi(x,y,z)}{\partial z} &= \lim_{h\to 0}\frac{\phi(x,y,z+h) - \phi(x,y,z)}{h}\\
&= \lim_{h\to 0}\frac{c_{2}(z+h)x^{4} - c_{2}zx^{4}}{h}\\
&= \lim_{h\to 0}\frac{c_{2}hx^{4}}{h}\\
&= c_{2}x^{4}.
\end{align}
A: While DMcMor's answer gives the rigorous reason why the partial derivative is $c_2x^4$, let me give you a more 'intuitive' treatment of how to compute partial derivatives quickly.
How I think about it when i have to compute something like $\frac{\partial}{\partial z}c_2x^4z$, is that I treat it as as if it was a function of $z$ only (since I'm interested in the partial derivative in $z$-direction), and treat the other variables (here $x^4$) as if they were constants. Then, computing the partial derivative here is nothing more than computing $f'(z)$, where $f(z) = a\cdot z$, which is of course just $f'(z) = a$. And this constant $a$ happens to be $a = c_2x^4$ in this case.
As I said, this is not as rigorous as using the definition, but there are situations in which you have to compute a lot of partial derivatives so it pays off to have a quick method.
