## Thomas' Calculus 13th Edition

We need to compute the Laplace's equation and prove that it equals to $0$. In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x=-2 e^{-2y} \sin 2x$; $f_y=-2 e^{-2y} \cos 2x$ Consider the Laplace's equation $\nabla^2 f =\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}$ $\nabla^2 f =-2 e^{-2y} \sin 2x-2 e^{-2y} \cos 2x +4 e^{-2y} \cos 2x=0$ Thus, the Laplace's equation is satisfied.