1. **Problem:** Given $z = \tan^{-1}\left(\frac{x}{y}\right)$, find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.\n\n2. **Formula and rules:** The derivative of $\tan^{-1}(u)$ with respect to $u$ is $\frac{1}{1+u^2}$. Use the chain rule for partial derivatives.\n\n3. **Step for $\frac{\partial z}{\partial x}$:**\nLet $u = \frac{x}{y}$. Then $z = \tan^{-1}(u)$.\n\n$$\frac{\partial z}{\partial x} = \frac{1}{1+u^2} \cdot \frac{\partial u}{\partial x}$$\n\nSince $u = \frac{x}{y}$, $\frac{\partial u}{\partial x} = \frac{1}{y}$.\n\nTherefore, $$\frac{\partial z}{\partial x} = \frac{1}{1+\left(\frac{x}{y}\right)^2} \cdot \frac{1}{y} = \frac{1}{1+\frac{x^2}{y^2}} \cdot \frac{1}{y} = \frac{1}{\frac{y^2+x^2}{y^2}} \cdot \frac{1}{y} = \frac{y^2}{y^2+x^2} \cdot \frac{1}{y} = \frac{y}{x^2 + y^2}.$$\n\n4. **Step for $\frac{\partial z}{\partial y}$:**\nAgain, $z = \tan^{-1}(u)$ with $u = \frac{x}{y}$.\n\n$$\frac{\partial z}{\partial y} = \frac{1}{1+u^2} \cdot \frac{\partial u}{\partial y}.$$\n\nCalculate $\frac{\partial u}{\partial y}$:\n$$u = \frac{x}{y} = x y^{-1} \implies \frac{\partial u}{\partial y} = x \cdot (-1) y^{-2} = -\frac{x}{y^2}.$$\n\nTherefore, $$\frac{\partial z}{\partial y} = \frac{1}{1+\left(\frac{x}{y}\right)^2} \cdot \left(-\frac{x}{y^2}\right) = \frac{1}{\frac{y^2+x^2}{y^2}} \cdot \left(-\frac{x}{y^2}\right) = \frac{y^2}{y^2+x^2} \cdot \left(-\frac{x}{y^2}\right) = -\frac{x}{x^2 + y^2}.$$\n\n**Final answers:**\n$$\frac{\partial z}{\partial x} = \frac{y}{x^2 + y^2}, \quad \frac{\partial z}{\partial y} = -\frac{x}{x^2 + y^2}.$$