Let. $X=e^{x-y}, \ Y=e^{y-z}, \ Z=e^{z-x} $

Then $u=f(X,Y,Z)$

$\therefore \dfrac {\partial u}{\partial x}= \dfrac {\partial u}{\partial X}\cdot \dfrac {\partial X}{\partial x} + \dfrac {\partial u}{\partial Y} \cdot \dfrac {\partial Y}{\partial x}+\dfrac {\partial u}{\partial Z} \cdot \dfrac {\partial Z}{\partial x}$

$\therefore \dfrac {\partial u}{\partial x} = \dfrac {\partial u}{\partial X} \cdot e^{x-y}(1) + \dfrac {\partial u}{\partial Y}\cdot (0)+ \dfrac {\partial u}{\partial Z}\cdot e^{z-x}(-1)$

$\therefore \dfrac {\partial u}{\partial x} = \dfrac {\partial u}{\partial X}\cdot e^{x-y}- \dfrac {\partial u}{\partial Z}\cdot e^{z-x}\ \cdots \ \cdots (I)$

$\therefore \dfrac {\partial u}{\partial y} = \dfrac {\partial u}{\partial X} \cdot \dfrac {\partial X}{\partial y} + \dfrac {\partial u}{\partial Y} \cdot \dfrac {\partial Y}{\partial y} + \dfrac {\partial u}{\partial Z}\cdot \dfrac {\partial Z}{\partial z}$

$ \therefore \dfrac {\partial u}{\partial y} = \dfrac {\partial u}{\partial X}\cdot e^{x-y}(-1)+ \dfrac {\partial u}{\partial Y}\cdot e^{y-z} + (1) \dfrac {\partial u}{\partial Z} \cdot (0)$

$\therefore \dfrac {\partial u}{\partial y}= \dfrac {\partial u} {\partial Y} \cdot e^{y - z} - \dfrac {\partial u}{\partial X} \cdot e^{x-y} \cdots \ \cdots (II)$

$\therefore \dfrac {\partial u}{\partial z} = \dfrac {\partial u}{\partial X}\cdot \dfrac {\partial X}{\partial z} + \dfrac {\partial u}{\partial Y} \cdot \dfrac {\partial Y}{\partial z} + \dfrac {\partial u}{\partial Z} \cdot \dfrac {\partial Z}{\partial z}$

$ \therefore \dfrac {\partial u}{\partial u} = \dfrac {\partial u}{\partial X}\cdot (0)+\dfrac {\partial u}{\partial Y}\cdot e^{y-z}(-1)+ \dfrac {\partial u}{\partial Z}\cdot e^{z-x} (1)$

$\therefore \dfrac {\partial u}{\partial z} = \dfrac {\partial u}{\partial Z} \cdot e^{z-x} - \dfrac {\partial u}{\partial Y} \cdot e^{y-z} \ \cdots \ \cdots (III)$

Adding (I), (II) and (III) we get

$\dfrac {\partial u} {\partial x} + \dfrac {\partial u}{\partial y} + \dfrac{\partial u} {\partial z} = 0$