The equation of a curve passing through the point $(0,1)$ be given by $y=\int x^2 . e^{x^3} d x$. If the equation of the curve be written in the form $x=f(y)$, then $f(y)=$

$\sqrt[3]{\log _e(3 y-2)}$

We have,

$y=\int x^2 e^{x^3} d x=\frac{1}{3} \int e^{x^3} d\left(x^3\right)=\frac{1}{3} e^{x^3}+C$

It passes through $(0,1)$. Therefore,

$1 =\frac{1}{3}+C \Rightarrow C=\frac{2}{3}$

∴  $y =\frac{1}{3} e^{x^3}+\frac{2}{3}$

$\Rightarrow 3 y=e^{x^3}+2$

$\Rightarrow e^{x^3}=3 y-2 \Rightarrow x^3=\log _e(3 y-2) \Rightarrow x=\sqrt[3]{\log _e(3 y-2)}$