Answer
$(\dfrac{\partial z }{\partial x})^2-(\dfrac{\partial z }{\partial y})^2 =\dfrac{\partial z }{\partial s}\dfrac{\partial z }{\partial t}$
Work Step by Step
Here, we have $\dfrac{\partial z }{\partial s} =\dfrac{\partial z }{\partial x} \times \dfrac{\partial x }{\partial s}+\dfrac{\partial z }{\partial y} \times \dfrac{\partial y }{\partial s}$
$\dfrac{\partial z }{\partial s} =\dfrac{\partial z }{\partial x} +\dfrac{\partial z }{\partial y} $
and $\dfrac{\partial z }{\partial t} =\dfrac{\partial z }{\partial x} -\dfrac{\partial z }{\partial y} $
$(\dfrac{\partial z }{\partial x})^2-(\dfrac{\partial z }{\partial y})^2 =\dfrac{1}{4}(\dfrac{\partial z }{\partial s})^2+\dfrac{1}{2}\dfrac{\partial z }{\partial s}\dfrac{\partial z }{\partial t}+\dfrac{1}{4} (\dfrac{\partial z }{\partial t})^2-[\dfrac{1}{4} (\dfrac{\partial z }{\partial s})^2-\dfrac{1}{2}\dfrac{\partial z }{\partial s}\dfrac{\partial z }{\partial t}+\dfrac{1}{4}(\dfrac{\partial z }{\partial t})^2]$
Therefore, we have $(\dfrac{\partial z }{\partial x})^2-(\dfrac{\partial z }{\partial y})^2 =\dfrac{\partial z }{\partial s}\dfrac{\partial z }{\partial t}$
