Answer
$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=e^{-2s}[ \dfrac{\partial^2 u}{\partial s^2}+\dfrac{\partial^2 u}{\partial t^2}]$
Work Step by Step
$\dfrac{\partial^2 u}{\partial s^2}=(u_{xx} \times x^2_s+u_x \times x_{ss})+(u_{xx}x^2_s+u_x \times x_{ss})+(u_{yy} \times y^2_s+u_y y_{ss})+(u_{xx} \times x^2_s+u_x x_{ss}) ; \\ \dfrac{\partial^2 u}{\partial t^2}=(u_{xx} \times x^2_t+u_x \times x_{tt})+(u_{xx}x^2_t+u_x \times x_{tt})+(u_{yy} \times y^2_t+u_y \times y_{tt})+(u_{xx}x^2_s+u_x x_{tt})$
Now, $\dfrac{\partial^2 u}{\partial s^2}+\dfrac{\partial^2 u}{\partial t^2}=e^{2s} \times (\dfrac{\partial^2 u}{\partial x^2} ) \times (\cos^2 t+\sin^2 t)+ e^{2s} \times ( \dfrac{\partial^2 u}{\partial y^2}) \times (\sin^2 t+\cos^2 t) $
This implies that
$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=(e^{-2s}) [ \dfrac{\partial^2 u}{\partial s^2}+\dfrac{\partial^2 u}{\partial t^2}] \bf({Verified})$
