1. **Problem Statement:** Given the function $$u = \frac{1}{\sqrt{x^2 + y^2}}$$, find the second partial derivatives $$\frac{\partial^2 u}{\partial x^2}$$ and $$\frac{\partial^2 u}{\partial x \partial y}$$. 2. **Recall the function:** $$u = (x^2 + y^2)^{-\frac{1}{2}}$$. 3. **First, find the first partial derivative with respect to x:** $$\frac{\partial u}{\partial x} = -\frac{1}{2} (x^2 + y^2)^{-\frac{3}{2}} \cdot 2x = -x (x^2 + y^2)^{-\frac{3}{2}}$$. 4. **Now, find the second partial derivative with respect to x:** Using the product and chain rules, $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(-x (x^2 + y^2)^{-\frac{3}{2}}\right)$$ $$= - (x^2 + y^2)^{-\frac{3}{2}} + x \cdot \frac{3}{2} (x^2 + y^2)^{-\frac{5}{2}} \cdot 2x$$ $$= - (x^2 + y^2)^{-\frac{3}{2}} + 3x^2 (x^2 + y^2)^{-\frac{5}{2}}$$ $$= \frac{- (x^2 + y^2) + 3x^2}{(x^2 + y^2)^{\frac{5}{2}}} = \frac{2x^2 - y^2}{(x^2 + y^2)^{\frac{5}{2}}}$$. 5. **Next, find the mixed second partial derivative $$\frac{\partial^2 u}{\partial x \partial y}$$:** Start with the first partial derivative with respect to x: $$\frac{\partial u}{\partial x} = -x (x^2 + y^2)^{-\frac{3}{2}}$$ Take the partial derivative with respect to y: $$\frac{\partial^2 u}{\partial x \partial y} = - \left[ (x^2 + y^2)^{-\frac{3}{2}} \cdot 0 + x \cdot \left(-\frac{3}{2}\right) (x^2 + y^2)^{-\frac{5}{2}} \cdot 2y \right]$$ $$= - x \cdot \left(-\frac{3}{2}\right) \cdot 2y (x^2 + y^2)^{-\frac{5}{2}} = 3xy (x^2 + y^2)^{-\frac{5}{2}}$$. **Final answers:** $$\frac{\partial^2 u}{\partial x^2} = \frac{2x^2 - y^2}{(x^2 + y^2)^{\frac{5}{2}}}$$ $$\frac{\partial^2 u}{\partial x \partial y} = \frac{3xy}{(x^2 + y^2)^{\frac{5}{2}}}$$