1. **State the problem:** Find the partial derivative of $z = x^2 - y^2 - \cos(xy)$ with respect to $y$, i.e., compute $\frac{\partial z}{\partial y}$. 2. **Recall the rules:** - The partial derivative with respect to $y$ treats $x$ as a constant. - Derivative of $x^2$ with respect to $y$ is 0 since $x$ is constant. - Derivative of $-y^2$ with respect to $y$ is $-2y$. - Derivative of $-\cos(xy)$ with respect to $y$ uses chain rule: derivative of $\cos(u)$ is $-\sin(u)$ times derivative of $u$. 3. **Apply the derivative:** $$\frac{\partial z}{\partial y} = 0 - 2y - \frac{\partial}{\partial y} \cos(xy)$$ 4. **Chain rule on $\cos(xy)$:** Let $u = xy$, then $$\frac{\partial}{\partial y} \cos(u) = -\sin(u) \cdot \frac{\partial u}{\partial y}$$ Since $u = xy$, $$\frac{\partial u}{\partial y} = x$$ 5. **Substitute back:** $$\frac{\partial}{\partial y} \cos(xy) = -\sin(xy) \cdot x = -x \sin(xy)$$ 6. **Combine all terms:** $$\frac{\partial z}{\partial y} = -2y - (-x \sin(xy)) = -2y + x \sin(xy)$$ **Final answer:** $$\boxed{-2y + x \sin(xy)}$$ This matches option (a).