1. **Problem statement:** Given the implicit equation $$x + y + z = \log z,$$ find the partial derivatives $$\frac{\partial z}{\partial x}$$ and $$\frac{\partial z}{\partial y}.$$ 2. **Formula and rules:** For implicit differentiation, treat $$z$$ as a function of $$x$$ and $$y$$, i.e., $$z = z(x,y)$$. Differentiate both sides of the equation with respect to $$x$$ and $$y$$ separately, applying the chain rule where necessary. 3. **Differentiate with respect to $$x$$:** $$\frac{\partial}{\partial x}(x + y + z) = \frac{\partial}{\partial x}(\log z)$$ Since $$y$$ is independent of $$x$$, $$\frac{\partial y}{\partial x} = 0$$. So, $$1 + 0 + \frac{\partial z}{\partial x} = \frac{1}{z} \frac{\partial z}{\partial x}$$ 4. **Rearrange to solve for $$\frac{\partial z}{\partial x}$$:** $$1 + \frac{\partial z}{\partial x} = \frac{1}{z} \frac{\partial z}{\partial x}$$ Multiply both sides by $$z$$: $$z + z \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}$$ Bring terms involving $$\frac{\partial z}{\partial x}$$ to one side: $$z = \frac{\partial z}{\partial x} - z \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}(1 - z)$$ Therefore, $$\frac{\partial z}{\partial x} = \frac{z}{1 - z}$$ 5. **Differentiate with respect to $$y$$:** $$\frac{\partial}{\partial y}(x + y + z) = \frac{\partial}{\partial y}(\log z)$$ Since $$x$$ is independent of $$y$$, $$\frac{\partial x}{\partial y} = 0$$. So, $$0 + 1 + \frac{\partial z}{\partial y} = \frac{1}{z} \frac{\partial z}{\partial y}$$ 6. **Rearrange to solve for $$\frac{\partial z}{\partial y}$$:** $$1 + \frac{\partial z}{\partial y} = \frac{1}{z} \frac{\partial z}{\partial y}$$ Multiply both sides by $$z$$: $$z + z \frac{\partial z}{\partial y} = \frac{\partial z}{\partial y}$$ Bring terms involving $$\frac{\partial z}{\partial y}$$ to one side: $$z = \frac{\partial z}{\partial y} - z \frac{\partial z}{\partial y} = \frac{\partial z}{\partial y}(1 - z)$$ Therefore, $$\frac{\partial z}{\partial y} = \frac{z}{1 - z}$$ **Final answers:** $$\boxed{\frac{\partial z}{\partial x} = \frac{z}{1 - z}, \quad \frac{\partial z}{\partial y} = \frac{z}{1 - z}}$$