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
$\dfrac{\partial u}{\partial s}=(e^s \cos t ) \times (\dfrac{\partial u}{\partial x})+( e^s \sin t) \times (\dfrac{\partial u}{\partial y})$
and $\dfrac{\partial u}{\partial t}=-(e^s \sin t) \times (\dfrac{\partial u}{\partial x}) + (e^s \cos t ) \times (\dfrac{\partial u}{\partial y})$
After solving , we get
$(\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2=e^{2s} \times [(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2]$
$(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$ ($\bf{Verified}$)
