[tex]\text{Given that,}\\\\\theta =\tan^{-1} \left(\dfrac y x \right)\\\\\\\dfrac{\partial \theta}{\partial x} = \dfrac{\partial }{\partial x} \tan^{-1} \left(\dfrac yx \right)\\ \\\\~~~~~~=\dfrac{1}{1 +\tfrac{y^2}{x^2}} \cdot y \dfrac{\partial }{\partial x} \left(\dfrac 1x \right)\\\\\\~~~~~~=\dfrac{yx^2}{x^2 +y^2} \cdot (-1)x^{-2}\\ \\\\~~~~~~=-\dfrac{yx^2}{x^2(x^2 +y^2)}\\\\\\~~~~~~=- \dfrac{y}{x^2 +y^2}\\\\\\[/tex]
[tex]\dfrac{\partial^2\theta}{ \partial x^2} = -\dfrac{\partial }{ \partial x} \left( \dfrac{y}{x^2+y^2} \right)\\\\\\~~~~~~=-y\dfrac{\partial}{\partial x} \left(\dfrac 1{x^2 +y^2} \right)\\\\\\~~~~~~=y (-1)(-1) (x^2 +y^2)^{-2} \dfrac{\partial }{\partial x}(x^2 +y^2)\\\\\\~~~~~~=\dfrac{y}{(x^2 +y^2)^2} \cdot (2x+0)\\\\\\~~~~~~=\dfrac{2xy}{(x^2 +y^2)^2}[/tex]
[tex]\text{Similarly,}~~\\ \\\dfrac{\partial \theta}{\partial y }= \dfrac{x}{x^2 +y^2}\\\\\\\dfrac{\partial^2 \theta}{\partial y^2 } = -\dfrac{2xy}{(x^2 +y^2)^2}\\\\\\\text{Now,}\\\\\dfrac{\partial^2 \theta}{\partial x^2 }+\dfrac{\partial^2 \theta}{\partial y^2 }\\\\\\=\dfrac{2xy}{(x^2 +y^2)^2} -\dfrac{2xy}{(x^2 +y^2)^2}\\\\\\=0[/tex]
