Solution: Note that if \[ f(z) = u(x,y)+ \iota v(x,y), \] then $f$ is analytic implies it satisfies the Cauchy-Riemann equations, that is, \begin{align*} u_x=v_y \text{ and } u_y = -v_x, \end{align*} where $u_x=\frac{\partial u}{\partial x}$.

Here given that \[ u(x,y) = e^{-x} \left( x \sin y-y \cos y \right). \] Using the Cauchy-Riemann equations, we have \begin{equation}\label{eq:v_y} v_y = u_x = e^{-x}\sin y-xe^{-x} \sin y+e^{-x} y\cos y \end{equation} \begin{equation}\label{eq:v_x} v_x = -u_y = -xe^{-x} \cos y+e^{-x}\cos y-ye^{-x} \sin y. \end{equation} So we need to find $v(x,y)$ such that $v_x$ and $v_y$ is given as above.

Integrating \eqref{eq:v_y} with respect to $y$ we obtain \begin{align*} v(x,y) & = \int \left( e^{-x}\sin y-xe^{-x} \sin y+e^{-x} y\cos y \right) \mathrm{d}y~+F(x) \\ & = \int e^{-x}\sin y\mathrm{d} y - \int xe^{-x} \sin y \mathrm{d} y + \int e^{-x} y\cos y \mathrm{d} y + F(x) \\ & = -e^{-x}\cos y + xe^{-x} \cos y + e^{-x}\int y \cos y\mathrm{d} y + F(x) \\ & = -e^{-x}\cos y + xe^{-x} \cos y + e^{-x}y\sin y+e^{-x}\cos y +F(x)\\ & = xe^{-x}\cos y+ye^{-x}\sin y+F(x), \end{align*} where $F(x)$ is some real function.

Now using the above substitute $v_x$ in the equation \eqref{eq:v_x}, \begin{align*} & v_x = e^{-x} \cos y-xe^{-x} \cos y-ye^{-x} \sin y + F^\prime (x) \\ \implies & F^\prime (x) = 0 \\ \implies & F(x) = c,~\text{for some constant $c$.} \end{align*} If we take $c=0$, then one possible $v(x,y)$ will be \[ \textcolor{blue}{\boxed{v(x,y) = ye^{-x}\sin y+xe^{-x}\cos y}} \]