1. **Problem statement:** Given the function $$u = \frac{x^2 + y^2}{x + y},$$ prove that $$\left(\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right)^2 = 4 \left(1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right).$$ 2. **Step 1: Compute partial derivatives.** Use the quotient rule for partial derivatives: if $$u = \frac{f}{g},$$ then $$\frac{\partial u}{\partial x} = \frac{g \frac{\partial f}{\partial x} - f \frac{\partial g}{\partial x}}{g^2}.$$ Here, $$f = x^2 + y^2$$ and $$g = x + y.$$ 3. **Calculate $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial g}{\partial x}$$:** $$\frac{\partial f}{\partial x} = 2x,$$ $$\frac{\partial g}{\partial x} = 1.$$ 4. **Calculate $$\frac{\partial u}{\partial x}$$:** $$\frac{\partial u}{\partial x} = \frac{(x + y)(2x) - (x^2 + y^2)(1)}{(x + y)^2} = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2}.$$ 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. **Calculate $$\frac{\partial u}{\partial y}$$ similarly:** $$\frac{\partial f}{\partial y} = 2y,$$ $$\frac{\partial g}{\partial y} = 1,$$ $$\frac{\partial u}{\partial y} = \frac{(x + y)(2y) - (x^2 + y^2)(1)}{(x + y)^2} = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2}.$$ 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. **Compute $$\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}$$:** $$= \frac{x^2 + 2xy - y^2}{(x + y)^2} - \frac{-x^2 + 2xy + y^2}{(x + y)^2} = \frac{(x^2 + 2xy - y^2) - (-x^2 + 2xy + y^2)}{(x + y)^2}.$$ Simplify numerator: $$x^2 + 2xy - y^2 + x^2 - 2xy - y^2 = 2x^2 - 2y^2 = 2(x^2 - y^2).$$ So, $$\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = \frac{2(x^2 - y^2)}{(x + y)^2}.$$ 7. **Square the difference:** $$\left(\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right)^2 = \left(\frac{2(x^2 - y^2)}{(x + y)^2}\right)^2 = \frac{4(x^2 - y^2)^2}{(x + y)^4}.$$ 8. **Compute $$1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}$$:** First sum the 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: $$x^2 + 2xy - y^2 - x^2 + 2xy + y^2 = 4xy.$$ So, $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{4xy}{(x + y)^2}.$$ Therefore, $$1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 1 - \frac{4xy}{(x + y)^2} = \frac{(x + y)^2 - 4xy}{(x + y)^2}.$$ 9. **Simplify numerator:** $$(x + y)^2 - 4xy = x^2 + 2xy + y^2 - 4xy = x^2 - 2xy + y^2 = (x - y)^2.$$ So, $$1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = \frac{(x - y)^2}{(x + y)^2}.$$ 10. **Multiply by 4:** $$4 \left(1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right) = \frac{4(x - y)^2}{(x + y)^2}.$$ 11. **Compare both sides:** Left side: $$\frac{4(x^2 - y^2)^2}{(x + y)^4} = \frac{4(x - y)^2 (x + y)^2}{(x + y)^4} = \frac{4(x - y)^2}{(x + y)^2}.$$ Right side: $$\frac{4(x - y)^2}{(x + y)^2}.$$ They are equal, so the identity is proven. **Final answer:** $$\left(\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right)^2 = 4 \left(1 - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right).$$