If \(u=f\left(\dfrac xy,\dfrac yz,\dfrac zx\right)\), prove that \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}+z\dfrac{\partial u}{\partial z}=0\).

Answer: Proved by the chain rule with \(r=\frac xy,\ s=\frac yz,\ t=\frac zx\).

Let \(r=\dfrac xy,\ s=\dfrac yz,\ t=\dfrac zx\), so \(u=f(r,s,t)\).

\(\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\cdot\dfrac1y+\dfrac{\partial u}{\partial t}\cdot\left(-\dfrac{z}{x^2}\right)\Rightarrow x\dfrac{\partial u}{\partial x}=r\dfrac{\partial u}{\partial r}-t\dfrac{\partial u}{\partial t}\) ...(1)

\(\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial r}\cdot\left(-\dfrac{x}{y^2}\right)+\dfrac{\partial u}{\partial s}\cdot\dfrac1z\Rightarrow y\dfrac{\partial u}{\partial y}=-r\dfrac{\partial u}{\partial r}+s\dfrac{\partial u}{\partial s}\) ...(2)

\(\dfrac{\partial u}{\partial z}=\dfrac{\partial u}{\partial s}\cdot\left(-\dfrac{y}{z^2}\right)+\dfrac{\partial u}{\partial t}\cdot\dfrac1x\Rightarrow z\dfrac{\partial u}{\partial z}=-s\dfrac{\partial u}{\partial s}+t\dfrac{\partial u}{\partial t}\) ...(3)

Adding (1), (2) and (3), every term cancels:

\[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=0.\]