1. **State the problem:** We need to find $\frac{dz}{dt}$ where $z = \sin(x) \cos(y)$, $x = t$, and $y = \frac{1}{t}$. 2. **Formula and rules:** Use the Chain Rule for multivariable functions: $$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$$ 3. **Calculate partial derivatives:** $$\frac{\partial z}{\partial x} = \cos(x) \cos(y)$$ $$\frac{\partial z}{\partial y} = -\sin(x) \sin(y)$$ 4. **Calculate derivatives of $x$ and $y$ with respect to $t$:** $$\frac{dx}{dt} = 1$$ $$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}$$ 5. **Substitute all into the Chain Rule formula:** $$\frac{dz}{dt} = \cos(x) \cos(y) \cdot 1 + (-\sin(x) \sin(y)) \cdot \left(-\frac{1}{t^2}\right)$$ 6. **Simplify:** $$\frac{dz}{dt} = \cos(t) \cos\left(\frac{1}{t}\right) + \sin(t) \sin\left(\frac{1}{t}\right) \frac{1}{t^2}$$ **Final answer:** $$\boxed{\frac{dz}{dt} = \cos(t) \cos\left(\frac{1}{t}\right) + \frac{\sin(t) \sin\left(\frac{1}{t}\right)}{t^2}}$$