Assume that \( w=f(u, v) \) where \( u=x+y \) and \( v=x-y \). Show that \[ \frac{\partial w}{\partial x} \frac{\partial w}{\partial y}=\left(\frac{\partial w}{\partial u}\right)^{2}-\left(\frac{\partial w}{\partial v}\right)^{2} \]

To show that \[ \frac{\partial w}{\partial x} \frac{\partial w}{\partial y} = \left(\frac{\partial w}{\partial u}\right)^{2} - \left(\frac{\partial w}{\partial v}\right)^{2}, \] we will use the chain rule to express \(\frac{\partial w}{\partial x}\) and \(\frac{\partial w}{\partial y}\) in terms of \(\frac{\partial w}{\partial u}\) and \(\frac{\partial w}{\partial v}\). First, we compute \(\frac{\partial w}{\partial x}\): Using the chain rule, we have: \[ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial w}{\partial v} \frac{\partial v}{\partial x}. \] Next, we need to find \(\frac{\partial u}{\partial x}\) and \(\frac{\partial v}{\partial x}\): \[ u = x + y \implies \frac{\partial u}{\partial x} = 1, \] \[ v = x - y \implies \frac{\partial v}{\partial x} = 1. \] Thus, we can substitute these into our expression for \(\frac{\partial w}{\partial x}\): \[ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} \cdot 1 + \frac{\partial w}{\partial v} \cdot 1 = \frac{\partial w}{\partial u} + \frac{\partial w}{\partial v}. \] Now, we compute \(\frac{\partial w}{\partial y}\): Using the chain rule again, we have: \[ \frac{\partial w}{\partial y} = \frac{\partial w}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial w}{\partial v} \frac{\partial v}{\partial y}. \] Next, we find \(\frac{\partial u}{\partial y}\) and \(\frac{\partial v}{\partial y}\): \[ u = x + y \implies \frac{\partial u}{\partial y} = 1, \] \[ v = x - y \implies \frac{\partial v}{\partial y} = -1. \] Substituting these into our expression for \(\frac{\partial w}{\partial y}\): \[ \frac{\partial w}{\partial y} = \frac{\partial w}{\partial u} \cdot 1 + \frac{\partial w}{\partial v} \cdot (-1) = \frac{\partial w}{\partial u} - \frac{\partial w}{\partial v}. \] Now we have: \[ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} + \frac{\partial w}{\partial v}, \] \[ \frac{\partial w}{\partial y} = \frac{\partial w}{\partial u} - \frac{\partial w}{\partial v}. \] Next, we compute the product \(\frac{\partial w}{\partial x} \frac{\partial w}{\partial y}\): \[ \frac{\partial w}{\partial x} \frac{\partial w}{\partial y} = \left(\frac{\partial w}{\partial u} + \frac{\partial w}{\partial v}\right) \left(\frac{\partial w}{\partial u} - \frac{\partial w}{\partial v}\right). \] Using the difference of squares, we can simplify this: \[ \frac{\partial w}{\partial x} \frac{\partial w}{\partial y} = \left(\frac{\partial w}{\partial u}\right)^{2} - \left(\frac{\partial w}{\partial v}\right)^{2}. \] Thus, we have shown that \[ \frac{\partial w}{\partial x} \frac{\partial w}{\partial y} = \left(\frac{\partial w}{\partial u}\right)^{2} - \left(\frac{\partial w}{\partial v}\right)^{2}. \] This completes the proof.