1. **State the problem:** Given $z = \log(x+y)$, with $x = 3u - v$ and $y = \frac{u^2}{2}$, find the partial derivatives $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}$ using the chain rule. 2. **Recall the chain rule for multivariable functions:** $$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}$$ $$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$$ 3. **Calculate the partial derivatives of $z$ with respect to $x$ and $y$:** Since $z = \log(x+y)$, $$\frac{\partial z}{\partial x} = \frac{1}{x+y}$$ $$\frac{\partial z}{\partial y} = \frac{1}{x+y}$$ 4. **Calculate the partial derivatives of $x$ and $y$ with respect to $u$ and $v$:** $$\frac{\partial x}{\partial u} = 3, \quad \frac{\partial x}{\partial v} = -1$$ $$\frac{\partial y}{\partial u} = \frac{d}{du} \left( \frac{u^2}{2} \right) = u, \quad \frac{\partial y}{\partial v} = 0$$ 5. **Substitute all into the chain rule formulas:** $$\frac{\partial z}{\partial u} = \frac{1}{x+y} \cdot 3 + \frac{1}{x+y} \cdot u = \frac{3+u}{x+y}$$ $$\frac{\partial z}{\partial v} = \frac{1}{x+y} \cdot (-1) + \frac{1}{x+y} \cdot 0 = \frac{-1}{x+y}$$ 6. **Rewrite $x+y$ in terms of $u$ and $v$:** $$x + y = (3u - v) + \frac{u^2}{2} = 3u - v + \frac{u^2}{2}$$ 7. **Final answers:** $$\boxed{\frac{\partial z}{\partial u} = \frac{3+u}{3u - v + \frac{u^2}{2}}}$$ $$\boxed{\frac{\partial z}{\partial v} = \frac{-1}{3u - v + \frac{u^2}{2}}}$$