Rules for partial derivatives Im trying to show some calculation rules for partial derivatives. 
Let $x,y,z$ be variables which are linked by  $f(x,y,z) = 0$. Given is a function $w(x,y)$ show that:


*

*$\displaystyle  \frac{\partial x}{\partial y} \Bigr|_z = \left(\frac{\partial y}{\partial x}\Bigr|_z\right)^{-1}$

*$\displaystyle (-1) = \left (\frac{\partial x}{\partial y}  \Bigr|_z\right) \left(\frac{\partial y}{\partial z}\Bigr|_x\right)\left( \frac{\partial z}{\partial x}\Bigr|_y\right)$

*$\displaystyle  \frac{\partial x}{\partial w} \Bigr|_z =  \left(\frac{\partial x}{\partial y}\Bigr|_z \right)\left( \frac{\partial y}{\partial w}\Bigr|_z\right)$


Now I'm confused by the wording of the exercise itself. What does "the variables are linked by $f(x,y,z) = 0$ mean? I guess $x(y,z), y(x,z), z(x,y)$ hence 1. isn't completely trivial? But how do I go on to show such relations? Do I use the definition of the differential quotient or how do I best show this? 
Im a physicist so please don't crucify me for not knowing this:)
Cheers in advance!
 A: Thermodynamics huh? :)


*

*If $f(x,y,z) = 0$ and $z = const$, then $\left .f(x,y,z)\right |_z = g(x,y) = 0$, which in turns means $x = h(y)$ (under the assumption $h$ exists). So if you differentiate it w.r. to $x$ and w.r. to $y$ you'll get
$$
\left .\frac {\partial x}{\partial y}\right|_z = h' \\
1 = \left .h'\frac {\partial y}{\partial x}\right |_z
$$
respectively, so
$$
\left .\frac {\partial x}{\partial y}\right|_z = \left (\left .\frac {\partial y}{\partial x}\right|_z \right)^{-1}
$$

*Since $f(x,y,z) = 0$ then e.g. $x = x(y,z)$. Find differential of this equation
$$
dx = \left .\frac {\partial x}{\partial y} \right |_z dy + \left .\frac {\partial x}{\partial z} \right |_y dz
$$
Now, recall that $x = const$, so $dx = 0$, hence $\left .dy \right|_x = \partial y_x$, and $\left . dz\right |_x = \partial z_x$
$$
\left .\frac {\partial x}{\partial y} \right |_z \partial y_x + \left .\frac {\partial x}{\partial z} \right |_y \partial z_x=0 \\
$$
Divide by $\partial z_x$
$$
\left .\frac {\partial x}{\partial y} \right |_z \frac {\partial y_x}{\partial z_x} + \left .\frac {\partial x}{\partial z} \right |_y = \left .\frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial z}\right |_x + \left .\frac {\partial x}{\partial z} \right |_y = 0
$$
Multiply it to $\left .\frac {\partial z}{\partial x} \right |_y$
$$
\left .\frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial z}\right |_x \left .\frac {\partial z}{\partial x} \right |_y+ 1 = 0
$$
Update

*Turned out, this is the easiest one. If $z=const$, then $w = w(x,y)$ and $y = y(x)$. Instead of proving 
$$
\left . \frac {\partial x}{\partial w} \right |_z = \left . \frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial w} \right |_z
$$
let's use the fact $\left . \frac {\partial x}{\partial w} \right |_z = \left (\left . \frac {\partial x}{\partial w} \right |_z \right)^{-1}$ and prove
$$
\left . \frac {\partial w}{\partial x} \right |_z = \left . \frac {\partial w}{\partial y} \right |_z \left . \frac {\partial y}{\partial x} \right |_z
$$
which is simple differentiation by chain rule.
