We are given (x = e^r \cos\theta) and (y = e^r \sin\theta), and (f(x,y)) is continuously differentiable. We want to show: [ \frac{\partial^2 f}{\partial r^2} + \frac{\partial^2 f}{\partial \theta^2} = (x^2 + y^2) \left(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\right). ] --- First, compute the first derivatives of (x) and (y) with respect to (r) and (\theta): [ \frac{\partial x}{\partial r} = e^r \cos\theta = x, \quad \frac{\partial x}{\partial \theta} = -e^r \sin\theta = -y, ] [ \frac{\partial y}{\partial r} = e^r \sin\theta = y, \quad \frac{\partial y}{\partial \theta} = e^r \cos\theta = x. ] --- By the chain rule, the first derivatives of (f) are: [ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r} = f_x \cdot x + f_y \cdot y, ] [ \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta} = f_x \cdot (-y) + f_y \cdot x = -f_x y + f_y x. ] --- Next, compute the second derivatives. Start with (\partial^2 f / \partial r^2): [ \frac{\partial^2 f}{\partial r^2} = \frac{\partial}{\partial r}(f_x x + f_y y). ] Apply the product and chain rules: [ \frac{\partial^2 f}{\partial r^2} = f_{xx} \left(\frac{\partial x}{\partial r}\right) x + f_x \frac{\partial x}{\partial