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 …