Solution: We recall that a function $f(x,y)$ is harmonic if the laplacian \[ \Delta f = f_{xx} + f_{yy} = 0. \] Let us compute the second partial derivatives of $u$. \begin{align*} u_x & = - e^{-x} \left( x \sin y - y \cos y \right) + e^{-x} \left( \sin y \right) \\ & = e^{-x} \left( \sin y-x\sin y + y\cos y \right) \\[2ex] u_{xx} & = e^{-x}\left( -\sin y + x\sin y- y\cos y - \sin y \right) \\ & = e^{-x}\left( -2\sin y + x\sin y- y\cos y\right) \\[2ex] u_y & = e^{-x}\left( x\cos y - \cos y + y\sin y \right) \\[2ex] u_{yy} & = e^{-x}\left( -x\sin y + \sin y + \sin y + y \cos y \right) \\ & = e^{-x} \left( 2\sin y- x \sin y + y \cos y \right) . \end{align*} Therefore, \begin{align*} \Delta u & = u_{xx} + u_{yy} \\ & = e^{-x}\left( \textcolor{red}{-2\sin y} + \textcolor{blue}{x\sin y} - \textcolor{teal}{y\cos y}\right) \\ & \kern 0.5cm + e^{-x} \left( \textcolor{red}{2\sin y} - \textcolor{blue}{x \sin y} + \textcolor{teal}{y \cos y} \right)\\ & = 0. \end{align*} Hence, we proved that $u$ is a harmonic function.

Now in order to find its harmonic conjugate $v$, we will use the Cauchy-Riemann equations. Recall that if $f = u + \iota v$ is analytic, then it must satisfy Cauchy-Riemann equations given by \begin{align*} u_x = v_y \text{ and } u_y = -v_x . \end{align*} Therefore, we have \begin{align*} & v_y = e^{-x} \left( \sin y-x\sin y + y\cos y \right) \\ \implies & v = \int \left[ e^{-x} \left( \sin y-x\sin y + y\cos y \right) \right] \mathrm{d} y \\ & \kern 0.2cm = e^{-x} \left( -\cos y + x \cos y + y \sin y + \cos y \right) + g(x)\\ & = \kern 0.2cm e^{-x} \left( x\cos y + y \sin y \right) + g(x), \end{align*} where $g(x)$ is an arbitrary function of $x$ only.

Now again from the Cauchy-Riemann equations, we have $v_x = - u_y$, which implies, \begin{align*} & e^{-x} \left( - x\cos y + \cos y - y\sin y \right) + g'(x) \\ & \kern 1cm = e^{-x}\left( -x\cos y + \cos y - y\sin y \right) \\ \implies & g'(x) = 0 \implies g(x) = k, \end{align*} where $k$ is a constant. Thus, \[ v(x,y) = e^{-x} \left( x \cos y + y \sin y \right) + k. \]