Answer: Let $u_1=e^{y-z}, u_2=e^{z-x}, u_3=e^{x-y}$. Then by chain rule,

$\dfrac{\partial u}{\partial x}=u_x=\dfrac{\partial u}{\partial u_1}\dfrac{\partial u_1}{\partial x}+\dfrac{\partial u}{\partial u_2}\dfrac{\partial u_2}{\partial x}+\dfrac{\partial u}{\partial u_3}\dfrac{\partial u_3}{\partial x}$

              $= \dfrac{\partial u}{\partial u_1}.0-\dfrac{\partial u}{\partial u_2}.e^{z-x}+\dfrac{\partial u}{\partial u_3}e^{x-y}.$

$\therefore \dfrac{\partial u}{\partial x}=-u_2 \dfrac{\partial u}{\partial u_2}+u_3\dfrac{\partial u}{\partial u_3}$...................................(1)

Similarly,

$\dfrac{\partial u}{\partial y}=u_y=\dfrac{\partial u}{\partial u_1}\dfrac{\partial u_1}{\partial y}+\dfrac{\partial u}{\partial u_2}\dfrac{\partial u_2}{\partial y}+\dfrac{\partial u}{\partial u_3}\dfrac{\partial u_3}{\partial y}$.

              $=\dfrac{\partial u}{\partial u_1}e^{y-z}+\dfrac{\partial u}{\partial u_2}.0-\dfrac{\partial u}{\partial u_3}e^{x-y}$

              $=u_1 \dfrac{\partial u}{\partial u_1}-u_3\dfrac{\partial u}{\partial u_3}$.................................(2)

$\dfrac{\partial u}{\partial z}=u_z=\dfrac{\partial u}{\partial u_1}\dfrac{\partial u_1}{\partial z}+\dfrac{\partial u}{\partial u_2}\dfrac{\partial u_2}{\partial z}+\dfrac{\partial u}{\partial u_3}\dfrac{\partial u_3}{\partial z}$

              $=-\dfrac{\partial u}{\partial u_1}e^{y-z}+\dfrac{\partial u}{\partial u_2}e^{z-x}+\dfrac{\partial u}{\partial u_3}.0$

              $=-u_1 \dfrac{\partial u}{\partial u_1}+u_2\dfrac{\partial u}{\partial u_2}$...............................(3)

Adding (1), (2) and (3), we get $u_x+u_y+u_z=0.$              Answer.