1. **State the problem:** Show that $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2u \log u$$ where $$\log u = \frac{x^3 + y^3}{3x + 4y}$$. 2. **Identify the function:** Let $$u = e^{\log u} = e^{\frac{x^3 + y^3}{3x + 4y}}$$. 3. **Calculate partial derivatives:** - Use the chain rule: $$\frac{\partial u}{\partial x} = u \frac{\partial}{\partial x} \left( \log u \right)$$ and $$\frac{\partial u}{\partial y} = u \frac{\partial}{\partial y} \left( \log u \right)$$. 4. **Compute $$\frac{\partial}{\partial x} \left( \log u \right)$$:** Let $$f = x^3 + y^3$$ and $$g = 3x + 4y$$, so $$\log u = \frac{f}{g}$$. Using quotient rule: $$\frac{\partial}{\partial x} \left( \frac{f}{g} \right) = \frac{g \frac{\partial f}{\partial x} - f \frac{\partial g}{\partial x}}{g^2}$$ Calculate derivatives: $$\frac{\partial f}{\partial x} = 3x^2$$ $$\frac{\partial g}{\partial x} = 3$$ So: $$\frac{\partial}{\partial x} \left( \log u \right) = \frac{(3x + 4y)(3x^2) - (x^3 + y^3)(3)}{(3x + 4y)^2}$$ 5. **Compute $$\frac{\partial}{\partial y} \left( \log u \right)$$:** Similarly, $$\frac{\partial}{\partial y} \left( \frac{f}{g} \right) = \frac{g \frac{\partial f}{\partial y} - f \frac{\partial g}{\partial y}}{g^2}$$ Calculate derivatives: $$\frac{\partial f}{\partial y} = 3y^2$$ $$\frac{\partial g}{\partial y} = 4$$ So: $$\frac{\partial}{\partial y} \left( \log u \right) = \frac{(3x + 4y)(3y^2) - (x^3 + y^3)(4)}{(3x + 4y)^2}$$ 6. **Substitute back into the expression:** $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x u \frac{\partial}{\partial x} \left( \log u \right) + y u \frac{\partial}{\partial y} \left( \log u \right) = u \frac{1}{(3x + 4y)^2} \left[ x \left( (3x + 4y)(3x^2) - 3(x^3 + y^3) \right) + y \left( (3x + 4y)(3y^2) - 4(x^3 + y^3) \right) \right]$$ 7. **Simplify the numerator:** Expand terms: $$x(3x + 4y)(3x^2) = 3x^3(3x + 4y) = 9x^4 + 12x^3 y$$ $$x \cdot (-3)(x^3 + y^3) = -3x^4 - 3x y^3$$ $$y(3x + 4y)(3y^2) = 3y^3(3x + 4y) = 9x y^3 + 12 y^4$$ $$y \cdot (-4)(x^3 + y^3) = -4 x^3 y - 4 y^4$$ Sum all: $$9x^4 + 12x^3 y - 3x^4 - 3x y^3 + 9x y^3 + 12 y^4 - 4 x^3 y - 4 y^4 = (9x^4 - 3x^4) + (12x^3 y - 4 x^3 y) + (-3x y^3 + 9x y^3) + (12 y^4 - 4 y^4) = 6x^4 + 8 x^3 y + 6 x y^3 + 8 y^4$$ 8. **Rewrite the numerator:** Group terms: $$6x^4 + 8 x^3 y + 6 x y^3 + 8 y^4 = 2(3x^4 + 4 x^3 y + 3 x y^3 + 4 y^4)$$ Notice that: $$3x^4 + 4 x^3 y + 3 x y^3 + 4 y^4 = (x^3 + y^3)(3x + 4y)$$ 9. **Therefore numerator is:** $$2 (x^3 + y^3)(3x + 4y)$$ 10. **Substitute back:** $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = u \frac{2 (x^3 + y^3)(3x + 4y)}{(3x + 4y)^2} = 2 u \frac{x^3 + y^3}{3x + 4y} = 2 u \log u$$ **Final answer:** $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2 u \log u$$ This completes the proof.