Chain rule for multi-variable functions So I have been studying the multi-variable chain rule.  Most importantly, and this is what I must have overlooked, is it's not always clear to me how to see which variables are functions of other variables, so that you know when to use the chain rule.  For example, if you have:
$$ x^2+y^2-z^2+2xy=1 $$
$$ x^3+y^3-5y=8 $$
In general, say we want to find $\frac{dz}{dt}$ but $z$ is a function of $x$, then we get:
$$ \frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} .$$  
And if $z$ is a function of both $y$ and $t$, we get:
$$ \frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt}$$  
In this case, we have two equations.  One involving all three variables $x,y,z$ and one involving just $x,y$.  Say we want to find $\frac{dz}{dx}$. What does this mean for this case?  How should we interpret this rule in general? 

You have 
$\displaystyle x^2+y^2-z^2+2y=1$  
$\displaystyle x^3+y^3-5y=8$  
You can take the derivative of each with respect to $x$ to get 
$\displaystyle 2x+2y\frac{dy}{dx}-2z\frac{dz}{dx}+2\frac{dy}{dx}=0$ 
$\displaystyle 3x^2+3y^2\frac{dy}{dx}-5\frac{dy}{dx}=0$
The second gives us  $\frac{dy}{dx}=\frac{3x^2}{3y^2-5}$ which you can substitute into the first to find $\frac{dz}{dx}$
Response to edit of the question:
The new first equation:  $\displaystyle x^2+y^2-z^2+2xy=1$  when we take the derivative by x gives $\displaystyle 2x+2y\frac{dy}{dx}-2z\frac{dz}{dx}+2x\frac{dy}{dx}+2y=0$, which can still be solved for dz/dx given the dy/dx of the second equation.
