$u = \tan^{-1}\Big(\dfrac{y}{x}\Big )\ \text{Partially differentiating this w.r.t x, treating y as constant} \ \dfrac{\partial u}{\partial x} = \dfrac{1}{1+ \Big(\dfrac{y}{x}\Big)^2} \times \dfrac{\partial }{\partial x} \Big(\dfrac{y}{x}\Big) \ \text{(using chain rule)} \ \quad \quad =\dfrac{x^2}{x^2+y^2 } \times \dfrac{-y}{x^2 } \ \quad \quad =\dfrac{-y}{x^2+y^2 } \ \text{Partially differentiating this again w.r.t .x,} \ \dfrac{\partial }{\partial x}\Big( \dfrac{\partial u}{\partial x}\Big) = \dfrac{\partial }{\partial x}\Big(\dfrac{-y}{x^2+y^2 } \Big) \ \dfrac{\partial^2 u}{\partial x^2} = -y \times \dfrac{-1}{(x^2+y^2)^2} \times \dfrac{\partial \Big(x^2+y^2 \Big)}{\partial x} \ \text{(using chain rule)} \ \dfrac{\partial^2 u}{\partial x^2} = \dfrac{2xy}{\Big(x^2+y^2 \Big)^2} \rightarrow [A]$ $u = \tan^{-1}\Big(\dfrac{y}{x}\Big) \ \text{Partially differentiating this w.r.t y, treating x as constant} \ \dfrac{\partial u}{\partial y} = \dfrac{1}{1+ \Big(\dfrac{y}{x}\Big)^2} \times \dfrac{\partial }{\partial y} (\dfrac{y}{x}\Big) \ \text{(using chain rule)} \ \quad \quad =\dfrac{x^2}{x^2+y^2 } \times \dfrac{1}{x } \ \quad \quad =\dfrac{x}{x^2+y^2 } \ \text{Partially differentiating this again w.r.t .y,} \ \dfrac{\partial }{\partial y}\Big( \dfrac{\partial u}{\partial y}\Big) = \dfrac{\partial }{\partial y}\Big(\dfrac{x}{x^2+y^2 } \Big) \ \dfrac{\partial^2 u}{\partial y^2} = x \times \dfrac{-1}{(x^2+y^2)^2} \times \dfrac{\partial (x^2+y^2)}{\partial y} \ \text{(using chain rule)} \ \dfrac{\partial^2 u}{\partial y^2} = \dfrac{-2xy}{(x^2+y^2)^2} \rightarrow [B]$ $\text{Adding [A] and [B],} $ $ \dfrac{\partial^2 u}{\partial x^2}+ \dfrac{\partial^2 u}{\partial y^2} = \dfrac{2xy}{(x^2+y^2)^2}+ \dfrac{-2xy}{(x^2+y^2)^2} \ { \dfrac{\partial^2 u}{\partial x^2}+ \dfrac{\partial^2 u}{\partial y^2} = 0 \quad }$