$\displaystyle Let\ V=e^x\left(x\sin y+y\cos y\right) \\[2ex] \displaystyle By\ Milne\ Thompson\ Method \\[2ex] $

$\displaystyle \therefore{}{\varnothing{}}_1=\frac{\partial{}v}{\partial{}x}=e^x.\left(x\sin y+y\cos y\right)+e^x(\sin y) \\[3ex] \ \ \ \ \varnothing{}_1={\ \varnothing{}}_{2(z,0)} \displaystyle =e^x\left(x\sin y+y\cos y+\sin y\right) \\[2ex] \displaystyle \therefore{}{\varnothing{}}_2=\frac{\partial{}v}{\partial{}y}=e^x\left(x\cos y+\cos y-y\sin y\right) \\[2ex] $

$= {\varnothing{}}_1\left(z,0\right) =ze^z+e^z\\[2ex] \displaystyle \therefore{}f\left(z\right)=\int\left[{\varnothing{}}_1\left(z,0\right)+i{\varnothing{}}_2\left(z,0\right)\right]dz+C\\[2ex] \displaystyle =\int\left[e^z(z+1)+i0\right]dz+C\\[2ex] \displaystyle =\left(z+1\right)e^z-\int e^zdz+C \\[2ex] \displaystyle =e^z\left(z+1\right)-e^z+C=ze^z+C\ which\ is\ the\ required\ function \\[2ex] \therefore{}\ It\ is\ Harmonic. $