1. The problem asks whether the function $f(x,y) = \cos\left(x - \frac{y}{3}\right)$ satisfies Laplace's equation. 2. Laplace's equation in two variables is given by: $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$$ 3. First, compute the first partial derivatives: $$\frac{\partial f}{\partial x} = -\sin\left(x - \frac{y}{3}\right)$$ $$\frac{\partial f}{\partial y} = -\sin\left(x - \frac{y}{3}\right) \cdot \left(-\frac{1}{3}\right) = \frac{1}{3} \sin\left(x - \frac{y}{3}\right)$$ 4. Next, compute the second partial derivatives: $$\frac{\partial^2 f}{\partial x^2} = -\cos\left(x - \frac{y}{3}\right)$$ $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{1}{3} \sin\left(x - \frac{y}{3}\right) \right) = \frac{1}{3} \cos\left(x - \frac{y}{3}\right) \cdot \left(-\frac{1}{3}\right) = -\frac{1}{9} \cos\left(x - \frac{y}{3}\right)$$ 5. Sum the second derivatives: $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = -\cos\left(x - \frac{y}{3}\right) - \frac{1}{9} \cos\left(x - \frac{y}{3}\right) = -\left(1 + \frac{1}{9}\right) \cos\left(x - \frac{y}{3}\right) = -\frac{10}{9} \cos\left(x - \frac{y}{3}\right)$$ 6. Since this sum is not zero for all $x,y$, the function does not satisfy Laplace's equation. **Final answer:** The function $f(x,y) = \cos\left(x - \frac{y}{3}\right)$ does not satisfy Laplace's equation because $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = -\frac{10}{9} \cos\left(x - \frac{y}{3}\right) \neq 0.$$