Why $\left(\frac{\partial x}{\partial y}\right)_z=-\left(\frac{\partial x}{\partial z}\right)_y \left(\frac{\partial z}{\partial y}\right)_x$? I was reading a statistical mechanics book, the author use: $\left(\frac{\partial x}{\partial y}\right)_z=-\left(\frac{\partial x}{\partial z}\right)_y \left(\frac{\partial z}{\partial y}\right)_x$ so I'm trying to prove this, but I'm stuck. I'm using the theorem:
$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}dz=0$
with $dz=\left(\frac{\partial z}{\partial x}\right)_y dx +\left(\frac{\partial z}{\partial y}\right)_xdy$
if $z$ is constant then $dz=0$ then $\left(\frac{\partial z}{\partial x}\right)_y dx =-\left(\frac{\partial z}{\partial y}\right)_xdy$ so I get:
$\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=0$
but this result doesn't lead me to anything
 A: In physics notation, you can write $dz=\left ( \frac{\partial z}{\partial x} \right )_y dx + \left ( \frac{\partial z}{\partial y} \right )_x dy=0$ (so you're holding $z$ fixed). Then solving for $dx$ you get $dx=-\frac{\left ( \frac{\partial z}{\partial y} \right )_x dy}{\left ( \frac{\partial z}{\partial x} \right )_y}$. Finally divide by $dy$, and interpret the result as holding $z$ fixed, giving $\left ( \frac{\partial x}{\partial y} \right )_z = -\frac{\left ( \frac{\partial z}{\partial y} \right )_x}{\left ( \frac{\partial z}{\partial x} \right )_y}$.
Finally to convert this to the form you wrote you apply the inverse function theorem in the denominator to get $\left ( \frac{\partial z}{\partial x} \right )_y = \left ( \left ( \frac{\partial x}{\partial z} \right )_y \right )^{-1}$.
In math notation you introduce a parametrized path $(x(t),y(t),f(x(t),y(t)))$ and check that $\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$, so if this is to be zero then $\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$. Then you use the inverse function theorem in the same way.
