1. **State the problem:** We are given the function $$f(x,y) = x^2 \arctan\left(\frac{y}{x}\right)$$ and need to find the second partial derivatives $$\frac{\partial^2 f}{\partial x^2}$$ and $$\frac{\partial^2 f}{\partial y^2}$$. 2. **Recall the formulas and rules:** - The first partial derivatives are found by differentiating with respect to one variable while treating the other as constant. - The second partial derivatives are the derivatives of the first partial derivatives. - Use the product rule and chain rule as needed. - Derivative of $$\arctan(u)$$ with respect to $$x$$ is $$\frac{1}{1+u^2} \cdot \frac{du}{dx}$$. 3. **Find the first partial derivative with respect to $$x$$:** $$f_x = \frac{\partial}{\partial x} \left(x^2 \arctan\left(\frac{y}{x}\right)\right)$$ Use product rule: $$f_x = 2x \arctan\left(\frac{y}{x}\right) + x^2 \cdot \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial x} \left(\frac{y}{x}\right)$$ 4. **Compute $$\frac{\partial}{\partial x} \left(\frac{y}{x}\right)$$:** Since $$y$$ is constant w.r.t. $$x$$, $$\frac{\partial}{\partial x} \left(\frac{y}{x}\right) = y \cdot \frac{\partial}{\partial x} \left(x^{-1}\right) = y \cdot (-x^{-2}) = -\frac{y}{x^2}$$ 5. **Substitute back:** $$f_x = 2x \arctan\left(\frac{y}{x}\right) + x^2 \cdot \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left(-\frac{y}{x^2}\right)$$ Simplify denominator: $$1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2}$$ So, $$f_x = 2x \arctan\left(\frac{y}{x}\right) - x^2 \cdot \frac{y}{x^2} \cdot \frac{x^2}{x^2 + y^2} = 2x \arctan\left(\frac{y}{x}\right) - \frac{y x^2}{x^2 + y^2} \cdot \frac{1}{x^2}$$ Cancel $$x^2$$: $$f_x = 2x \arctan\left(\frac{y}{x}\right) - \frac{y}{x^2 + y^2}$$ 6. **Find the second partial derivative with respect to $$x$$:** $$f_{xx} = \frac{\partial}{\partial x} f_x = \frac{\partial}{\partial x} \left(2x \arctan\left(\frac{y}{x}\right) - \frac{y}{x^2 + y^2}\right)$$ 7. **Differentiate the first term:** Use product rule: $$\frac{\partial}{\partial x} \left(2x \arctan\left(\frac{y}{x}\right)\right) = 2 \arctan\left(\frac{y}{x}\right) + 2x \cdot \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right)$$ Simplify denominator as before: $$= 2 \arctan\left(\frac{y}{x}\right) - \frac{2xy}{x^2 + y^2}$$ 8. **Differentiate the second term:** $$\frac{\partial}{\partial x} \left(- \frac{y}{x^2 + y^2}\right) = -y \cdot \frac{\partial}{\partial x} \left(\frac{1}{x^2 + y^2}\right) = -y \cdot \left(- \frac{2x}{(x^2 + y^2)^2}\right) = \frac{2xy}{(x^2 + y^2)^2}$$ 9. **Combine results:** $$f_{xx} = 2 \arctan\left(\frac{y}{x}\right) - \frac{2xy}{x^2 + y^2} + \frac{2xy}{(x^2 + y^2)^2}$$ 10. **Find the first partial derivative with respect to $$y$$:** $$f_y = \frac{\partial}{\partial y} \left(x^2 \arctan\left(\frac{y}{x}\right)\right) = x^2 \cdot \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial y} \left(\frac{y}{x}\right)$$ Since $$x$$ is constant w.r.t. $$y$$, $$\frac{\partial}{\partial y} \left(\frac{y}{x}\right) = \frac{1}{x}$$ 11. **Substitute:** $$f_y = x^2 \cdot \frac{1}{1 + \frac{y^2}{x^2}} \cdot \frac{1}{x} = x^2 \cdot \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \frac{1}{x} = x^2 \cdot \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x}$$ Simplify: $$f_y = \frac{x^3}{x^2 + y^2}$$ 12. **Find the second partial derivative with respect to $$y$$:** $$f_{yy} = \frac{\partial}{\partial y} f_y = \frac{\partial}{\partial y} \left(\frac{x^3}{x^2 + y^2}\right)$$ Treat $$x^3$$ as constant: $$f_{yy} = x^3 \cdot \frac{\partial}{\partial y} \left(\frac{1}{x^2 + y^2}\right) = x^3 \cdot \left(- \frac{2y}{(x^2 + y^2)^2}\right) = - \frac{2x^3 y}{(x^2 + y^2)^2}$$ **Final answers:** $$\boxed{\frac{\partial^2 f}{\partial x^2} = 2 \arctan\left(\frac{y}{x}\right) - \frac{2xy}{x^2 + y^2} + \frac{2xy}{(x^2 + y^2)^2}}$$ $$\boxed{\frac{\partial^2 f}{\partial y^2} = - \frac{2x^3 y}{(x^2 + y^2)^2}}$$