Given u=f(x-y, y-z, z-x)

Let x-y=a,y-z=b.z-x=c then u=f(a,b,c)

• Therefore differentiating a,b,c with respect to x we have

   $\dfrac {\partial a}{\partial x}=1$        $\dfrac{\partial b}{\partial x}=0$        $\dfrac{\partial c}{\partial x}=-1$

   

• Similarily we have

$\dfrac{\partial a}{\partial y}=-1$      $\dfrac{\partial b}{\partial y}=1$         $\dfrac{\partial c}{\partial y}=0$

$\dfrac{\partial a}{\partial z}=0$         $\dfrac{\partial b}{\partial z}=-1$      $\dfrac{\partial c}{\partial z}=1$

• Therefore$\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial a}\times\dfrac{\partial a}{\partial x}+\dfrac{\partial u}{\partial b}\times\dfrac{\partial b}{\partial x}+\dfrac{\partial u}{\partial c}\times\dfrac{\partial c}{\partial x}$

$\dfrac {\partial u}{\partial x}=\dfrac {\partial u}{\partial a}\times1+\dfrac {\partial u}{\partial b}\times0+\dfrac {\partial u}{\partial c}\times(-1)$$=\dfrac {\partial u}{\partial a}-\dfrac {\partial u}{\partial c}$...........1   * Similarily$\dfrac {\partial u}{\partial y}=\dfrac {\partial u}{\partial a}\times\dfrac {\partial a}{\partial y}+\dfrac {\partial u}{\partial b}\times\dfrac {\partial b}{\partial y}+\dfrac {\partial u}{\partial c}\times\dfrac {\partial c}{\partial y}$ $\dfrac {\partial u}{\partial y}=\dfrac {\partial u}{\partial a}\times(-1)+\dfrac {\partial u}{\partial b}\times(1)+\dfrac {\partial u}{\partial c}\times0$$=\dfrac {\partial u}{\partial b}-\dfrac {\partial u}{\partial a}$.........2   * Similarily$\dfrac {\partial u}{\partial z}=\dfrac {\partial u}{\partial a}\times\dfrac {\partial a}{\partial z}+\dfrac {\partial u}{\partial b}\times\dfrac {\partial b}{\partial z}+\dfrac {\partial u}{\partial c}\times\dfrac {\partial c}{\partial z}$ $\dfrac {\partial u}{\partial z}=\dfrac {\partial u}{\partial a}\times0+\dfrac {\partial u}{\partial b}\times(-1)+\dfrac {\partial u}{\partial c}\times(1)$$=\dfrac {\partial u}{\partial c}-\dfrac {\partial u}{\partial b}$........3   * Adding 1 ,2 and 3 we$\dfrac {\partial u}{\partial x}+ \dfrac {\partial u}{\partial y} + \dfrac {\partial u}{\partial z} = 0$get