1. **State the problem:** We are given the variables $x$ and $y$ defined in terms of $u$ and $v$ as: $$x = \frac{u}{v}, \quad y = \frac{1}{v - u}$$ We need to find the Jacobian determinant $\frac{\partial(x,y)}{\partial(u,v)}$. 2. **Recall the Jacobian formula:** The Jacobian determinant for the transformation from $(u,v)$ to $(x,y)$ is: $$J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}$$ 3. **Calculate partial derivatives:** - For $x = \frac{u}{v}$: $$\frac{\partial x}{\partial u} = \frac{1}{v}$$ $$\frac{\partial x}{\partial v} = \frac{\partial}{\partial v} \left( \frac{u}{v} \right) = u \cdot \frac{\partial}{\partial v} \left( \frac{1}{v} \right) = u \cdot \left(-\frac{1}{v^2}\right) = -\frac{u}{v^2}$$ - For $y = \frac{1}{v - u}$: $$\frac{\partial y}{\partial u} = \frac{\partial}{\partial u} \left( (v - u)^{-1} \right) = -1 \cdot (v - u)^{-2} \cdot (-1) = \frac{1}{(v - u)^2}$$ $$\frac{\partial y}{\partial v} = \frac{\partial}{\partial v} \left( (v - u)^{-1} \right) = -1 \cdot (v - u)^{-2} \cdot 1 = -\frac{1}{(v - u)^2}$$ 4. **Substitute into the Jacobian formula:** $$J = \frac{1}{v} \cdot \left(-\frac{1}{(v - u)^2}\right) - \left(-\frac{u}{v^2}\right) \cdot \frac{1}{(v - u)^2}$$ 5. **Simplify the expression:** $$J = -\frac{1}{v (v - u)^2} + \frac{u}{v^2 (v - u)^2}$$ Combine the terms over the common denominator $v^2 (v - u)^2$: $$J = \frac{-v + u}{v^2 (v - u)^2} = \frac{u - v}{v^2 (v - u)^2}$$ 6. **Final answer:** The Jacobian determinant is: $$\boxed{\frac{u - v}{v^2 (v - u)^2}}$$ This corresponds to option (a).