1. **State the problem:** We are given the function $z = x^2 - y^2 - \cos(xy)$ and need to find the partial derivative of $z$ with respect to $y$, denoted as $\frac{\partial z}{\partial y}$. 2. **Recall the rules:** - The partial derivative with respect to $y$ treats $x$ as a constant. - The derivative of $y^2$ with respect to $y$ is $2y$. - The derivative of $\cos(u)$ with respect to $y$ is $-\sin(u) \cdot \frac{du}{dy}$ by the chain rule. 3. **Apply the derivative term-by-term:** - $\frac{\partial}{\partial y}(x^2) = 0$ since $x^2$ is constant with respect to $y$. - $\frac{\partial}{\partial y}(-y^2) = -2y$. - For $-\cos(xy)$, let $u = xy$, then $\frac{du}{dy} = x$. So, $\frac{\partial}{\partial y}(-\cos(xy)) = -(-\sin(xy) \cdot x) = x \sin(xy)$. 4. **Combine all parts:** $$\frac{\partial z}{\partial y} = 0 - 2y + x \sin(xy) = -2y + x \sin(xy)$$ 5. **Final answer:** The correct option is (a) $-2y + x \sin(xy)$.