Answer
$\dfrac{\partial z }{\partial x} +\dfrac{\partial z }{\partial y}=0$
Work Step by Step
Consider $a=x-y$
Here, we have $\dfrac{\partial z }{\partial x} =\dfrac{\partial z }{\partial a} \times \dfrac{\partial a }{\partial x}$
$\dfrac{\partial z }{\partial x} =\dfrac{\partial z }{\partial a} \times (1) $
$\dfrac{\partial z }{\partial x} =\dfrac{\partial z }{\partial a} \times $
Now, $\dfrac{\partial z }{\partial y} =\dfrac{\partial z }{\partial a} \times \dfrac{\partial a }{\partial y}$
$\dfrac{\partial z }{\partial y} =\dfrac{\partial z }{\partial a} \times (-1) $
$\dfrac{\partial z }{\partial y} =-\dfrac{\partial z }{\partial a}$
Therefore, we have $\dfrac{\partial z }{\partial x} +\dfrac{\partial z }{\partial y}=0$
