$F(x,y,z)=0$ differentiation

Question

Suppose $F:\mathbb R^3\rightarrow\mathbb R$ is $C^1$, and $F(x,y,z)=0$. $z = f (x, y), x = g(y, z)$, and $y = h(x,z)$.

Show that $\frac{\partial z~\partial x~\partial y}{\partial x ~\partial y~\partial z}=-1$.

I know for $f(x,y)=0$, we get $\frac{dy}{dx}=-\frac{\frac{df}{dx}}{\frac{df}{dy}}$.

Answer 1

Differentiate the equation $F(x,y,f(x,y))= 0$ with respect to $x$ to get \begin{equation*} F_x + F_z\frac{\partial f}{\partial x} = 0. \end{equation*} Differentiate the equation $F(x, h(x,z), z) = 0$ to with respect to $z$ to get \begin{equation*} F_y\frac{\partial h}{\partial z} + F_z=0. \end{equation*} Differentiate the equation $F(g(y,z), y, z)= 0$ to wth respect to $y$ to get \begin{equation*} F_x\frac{\partial g}{\partial y} + F_y = 0. \end{equation*} Using the three computations above gives \begin{equation*} \frac{\partial f}{\partial x} \frac{\partial g}{\partial y}\frac{\partial h}{\partial z} = \left(-\frac{F_x}{F_z}\right)\left(-\frac{F_y}{F_x}\right)\left(- \frac{F_z}{F_y}\right)= -1. \end{equation*}