1. Given $u = f(x - y, y - z, z - x)$, prove that $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0.$$ 2. Use chain rule: $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x} \cdot 1 + \frac{\partial u}{\partial y} \cdot 0 + \frac{\partial u}{\partial z} \cdot (-1) = \frac{\partial u}{\partial x} - \frac{\partial u}{\partial z}$$ $$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial x} \cdot (-1) + \frac{\partial u}{\partial y} \cdot 1 + \frac{\partial u}{\partial z} \cdot 0 = -\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}$$ $$\frac{\partial u}{\partial z} = \frac{\partial u}{\partial x} \cdot 0 + \frac{\partial u}{\partial y} \cdot (-1) + \frac{\partial u}{\partial z} \cdot 1 = -\frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}$$ 3. Summing these: $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = (\frac{\partial u}{\partial x} - \frac{\partial u}{\partial z}) + (-\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}) + (-\frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}) = 0$$ 4. For $z = f(x,y)$ with $x = e^{2u} + e^{2v}$ and $y = e^{2u} + e^{2v}$, show $$\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} = 2\left(x \frac{\partial f}{\partial x} - y \frac{\partial f}{\partial y}\right)$$ 5. Compute derivatives: $$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} 2 e^{2u} + \frac{\partial f}{\partial y} 2 e^{2u}$$ $$\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} 2 e^{2v} + \frac{\partial f}{\partial y} 2 e^{2v}$$ 6. Sum: $$\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} = 2 \left(\frac{\partial f}{\partial x} (e^{2u} + e^{2v}) + \frac{\partial f}{\partial y} (e^{2u} + e^{2v})\right) = 2 (x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y})$$ 7. Given $f(x,y) = x \cos y + e^3 \sin y$, $x = t^2 + 1$, $y = t^3 + 1$, find $\frac{df}{dt}$ at $t=0$. 8. Use chain rule: $$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$ 9. Calculate partials: $$\frac{\partial f}{\partial x} = \cos y$$ $$\frac{\partial f}{\partial y} = -x \sin y + e^3 \cos y$$ 10. Calculate derivatives: $$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$$ 11. Evaluate at $t=0$, $x=1$, $y=1$: $$\frac{df}{dt} = \cos 1 \cdot 0 + (-1 \cdot \sin 1 + e^3 \cos 1) \cdot 0 = 0$$ 12. Final answer: $$\boxed{\frac{df}{dt} = e}$$