Let's solve this step by step manually. We are asked to compute the mixed partial derivative of the function $$ f(x, y) = x^2 \tan^{-1}\frac{y}{x} - y \tan^{-1}\frac{x}{y}, \quad f(0,0)=0, $$ and show that $$ \frac{\partial^2 f}{\partial x \partial y} = \frac{x^2 - y^2}{x^2 + y^2}. $$ --- ### Step 1: Compute $\frac{\partial f}{\partial x}$ We have $$ f(x,y) = x^2 \tan^{-1}\frac{y}{x} - y \tan^{-1}\frac{x}{y}. $$ #### First term: $x^2 \tan^{-1}(y/x)$ Let $u = y/x \Rightarrow \tan^{-1}(y/x) = \tan^{-1}(u)$. $$ \frac{\partial}{\partial x} [x^2 \tan^{-1}(y/x)] = 2x \tan^{-1}(y/x) + x^2 \cdot \frac{d}{dx} \tan^{-1}(y/x) $$ We know $\frac{d}{dx} \tan^{-1}(y/x) = \frac{1}{1+(y/x)^2} \cdot \frac{d}{dx} (y/x) = \frac{1}{1+(y^2/x^2)} \cdot (-y/x^2) = \frac{-y/x^2}{1 + y^2/x^2} = \frac{-y}{x^2 + y^2}$. So, $$ \frac{\partial}{\partial x} [x^2 \tan^{-1}(y/x)] = 2x \tan^{-1}(y/x) + x^2 \cdot \left(-\frac{y}{x^2 + y^2}\right) = 2x \tan^{-1}(y/x) - \frac{x^2 y}{x^2 + y^2}. $$ --- #### Second term: $- y \tan^{-1}(x/y)$ $$ \frac{\partial}{\partial x} [-y \tan^{-1}(x/y)] = -y \cdot \frac{1}{1+(x/y)^2} \cdot \frac{\partial}{\partial x} (x/y) = -y \cdot \frac{1}{1 + x^2/y^2} \cdot (1/y) $$ $$ = -y \cdot \frac{1}{(y^2 + x^2)/y^2} \cdot \frac{1}{y} = -y \cdot \frac{y^2}{x^2 + y^2} \cdot \frac{1}{y} = - \frac{y^2}{x^2 + y^2}. $$ --- #### Combine both terms $$ \frac{\partial f}{\partial x} = 2x \tan^{-1}(y/x) - \frac{x^2 y}{x^2 + y^2} - \frac{y^2}{x^2 + y^2} = 2x \tan^{-1}(y/x) - \frac{x^2 y + y^2}{x^2 + y^2}. $$ Notice $-\frac{x^2 y + y^2}{x^2 + y^2} = -\frac{y(x^2 + y)}{x^2 + y^2} $. We'll simplify later. --- ### Step 2: Compute $\frac{\partial^2 f}{\partial x \partial y}$ We now differentiate $\frac{\partial f}{\partial x}$ w\.r.t $y$. $$ \frac{\partial f}{\partial x} = 2x \tan^{-1}(y/x) - \frac{x^2 y + y^2}{x^2 + y^2} = 2x \tan^{-1}(y/x) - \frac{y(x^2 + y)}{x^2 + y^2} $$ #### First term: