Find both first partial derivatives. \[ z=\cos (3 x y) \]

To find the first partial derivatives of the function \( z = \cos(3xy) \), we will differentiate \( z \) with respect to \( x \) and \( y \). 1. The partial derivative with respect to \( x \) is given by applying the chain rule: \[ \frac{\partial z}{\partial x} = -\sin(3xy) \cdot \frac{\partial}{\partial x}(3xy) = -\sin(3xy) \cdot (3y) = -3y \sin(3xy). \] 2. The partial derivative with respect to \( y \) is similarly found: \[ \frac{\partial z}{\partial y} = -\sin(3xy) \cdot \frac{\partial}{\partial y}(3xy) = -\sin(3xy) \cdot (3x) = -3x \sin(3xy). \] So the first partial derivatives are: \[ \frac{\partial z}{\partial x} = -3y \sin(3xy), \] \[ \frac{\partial z}{\partial y} = -3x \sin(3xy). \]