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
Here, we have $\dfrac{\partial^2 u}{\partial s^2}=(u_{xx}x^2_s+u_x x_{ss})+(u_{xx}x^2_s+u_x x_{ss})+(u_{yy}y^2_s+u_y y_{ss})+(u_{xx}x^2_s+u_x x_{ss})$
and $\dfrac{\partial^2 u}{\partial t^2}=(u_{xx}x^2_t+u_x x_{tt})+(u_{xx}x^2_t+u_x x_{tt})+(u_{yy}y^2_t+u_y y_{tt})+(u_{xx}x^2_s+u_x x_{tt})$
Now, $\dfrac{\partial^2 u}{\partial s^2}+\dfrac{\partial^2 u}{\partial t^2}=\dfrac{\partial^2 u}{\partial x^2} e^{2s} (\cos^2 t+\sin^2 t)+ e^{2s} \dfrac{\partial^2 u}{\partial y^2}(\sin^2 t+\cos^2 t) $
Since, $\sin^2 t+\cos^2 t=1$
Therefore, it has been verified that $\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=e^{-2s}[ \dfrac{\partial^2 u}{\partial s^2}+\dfrac{\partial^2 u}{\partial t^2}]$
