1. **Problem Statement:** Find the value of $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}$ where $u = \frac{x^2 + y^2}{x + y}$. 2. **Formula and Rules:** We use partial differentiation rules. For a function $u = \frac{f(x,y)}{g(x,y)}$, the partial derivative with respect to $x$ is given by the quotient rule: $$\frac{\partial u}{\partial x} = \frac{g \frac{\partial f}{\partial x} - f \frac{\partial g}{\partial x}}{g^2}$$ Similarly for $y$. 3. **Calculate partial derivatives:** Let $f = x^2 + y^2$ and $g = x + y$. Calculate: $$\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial g}{\partial x} = 1$$ $$\frac{\partial f}{\partial y} = 2y, \quad \frac{\partial g}{\partial y} = 1$$ 4. **Compute $\frac{\partial u}{\partial x}$:** $$\frac{\partial u}{\partial x} = \frac{(x + y)(2x) - (x^2 + y^2)(1)}{(x + y)^2} = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2}$$ Simplify numerator: $$2x^2 + 2xy - x^2 - y^2 = x^2 + 2xy - y^2$$ So: $$\frac{\partial u}{\partial x} = \frac{x^2 + 2xy - y^2}{(x + y)^2}$$ 5. **Compute $\frac{\partial u}{\partial y}$:** $$\frac{\partial u}{\partial y} = \frac{(x + y)(2y) - (x^2 + y^2)(1)}{(x + y)^2} = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2}$$ Simplify numerator: $$2xy + 2y^2 - x^2 - y^2 = -x^2 + 2xy + y^2$$ So: $$\frac{\partial u}{\partial y} = \frac{-x^2 + 2xy + y^2}{(x + y)^2}$$ 6. **Add the two partial derivatives:** $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{x^2 + 2xy - y^2}{(x + y)^2} + \frac{-x^2 + 2xy + y^2}{(x + y)^2} = \frac{(x^2 + 2xy - y^2) + (-x^2 + 2xy + y^2)}{(x + y)^2}$$ Simplify numerator: $$x^2 - x^2 + 2xy + 2xy - y^2 + y^2 = 4xy$$ Therefore: $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{4xy}{(x + y)^2}$$ **Final answer:** $$\boxed{\frac{4xy}{(x + y)^2}}$$