1. **State the problem:** We need to find the partial derivative of the function $$Z = x^3 y^2 - \frac{y}{x^2} + \frac{1}{y}$$ with respect to $$x$$. 2. **Recall the rules:** When taking the partial derivative with respect to $$x$$, treat $$y$$ as a constant. 3. **Differentiate each term:** - For $$x^3 y^2$$, since $$y^2$$ is constant, $$\frac{\partial}{\partial x}(x^3 y^2) = y^2 \cdot \frac{\partial}{\partial x}(x^3) = y^2 \cdot 3x^2 = 3x^2 y^2$$. - For $$- \frac{y}{x^2}$$, rewrite as $$- y x^{-2}$$. Treating $$y$$ as constant, $$\frac{\partial}{\partial x}(- y x^{-2}) = - y \cdot (-2) x^{-3} = 2 y x^{-3}$$. - For $$\frac{1}{y}$$, since it does not depend on $$x$$, its partial derivative with respect to $$x$$ is $$0$$. 4. **Combine the results:** $$\frac{\partial Z}{\partial x} = 3x^2 y^2 + 2 y x^{-3} + 0 = 3x^2 y^2 + \frac{2 y}{x^3}$$. **Final answer:** $$\boxed{\frac{\partial Z}{\partial x} = 3x^2 y^2 + \frac{2 y}{x^3}}$$