Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 22

$$ y =\frac{3}{3x-x^3+3c}.$$

We have $$ y'= y^2(1-x^2)\Longrightarrow \frac{dy}{y^2}=(1-x^2)dx\\ \Longrightarrow\int \frac{dy}{y^2}=\int (1-x^2)dx+c\\ \Longrightarrow -y^{-1}=x-\frac{1}{3}x^3+c \Longrightarrow y =\frac{3}{3x-x^3+3c}.$$ Hence, the general solution is $$ y =\frac{3}{3x-x^3+3c}.$$