Chapter 1 - First-Order Differential Equations - 1.3 The Geometry of First-Order Differential Equations - Problems - Page 32: 14

See answer below

We are given: $y'=\frac{x}{x^2+1}(y^2-9)$ $\frac{\partial f}{\partial y}=\frac{2xy}{x^2+1}$ These equations are continuous along the entirety of the (x,y) plane, thus: $y'=\frac{x}{x^2+1}(y^2-9)$ $\rightarrow y(0)=3$ By the Existence and Uniqueness Theorem, the differential has a unique solution. Since $y(x) = 3$ $\rightarrow y'=\frac{x}{x^2+1}(3^2-9)=0$ $y(x)=3$ is the only solution to the given initial-value problem.