Answer
\[y=\frac{3}{2-x^3}\]
Work Step by Step
Slope of tangent to a curve at ($x,y$) is given by $\frac{dy}{dx}$ at ($x,y$)
\[\frac{dy}{dx}=x^2y^2\]
Seperating variables
\[\frac{dy}{y^2}=x^2 dx\]
Integrating,
\[\int y^{-2}dy=\int x^2 dx+C\]
Where C is constant of integration
$\frac{y^{-1}}{-1}=\frac{x^3}{3}+C$ ____(1)
Curve passes through $(-1,1)$
\[-1=\frac{-1}{3}+C \Rightarrow C=\frac{-2}{3}\]
From (1)
\[\frac{-1}{y}=\frac{x^3-2}{3} \Rightarrow y=\frac{3}{2-x^3}\]
Hence, equation of curve is $y=\frac{3}{2-x^3}$
