Answer
$(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$
Work Step by Step
Consider $(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$
Now, we have $\dfrac{\partial u}{\partial s}=e^s \cos t \dfrac{\partial u}{\partial x}+ e^s \sin t \dfrac{\partial u}{\partial y} ; \\ \dfrac{\partial u}{\partial t}=-e^s \sin t \dfrac{\partial u}{\partial x}+ e^s \cos t \dfrac{\partial u}{\partial y} $
and $(\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2=e^{2s}[(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2]$
So, it has been verified that $(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$
