#### Answer

(a) $y^{3} d y =9 x^{2} d x$
(b) $\frac{y^{4}}{4} =3 x^{3}+C$
(c) $ y= (12 x^{3}+C)^{1/4}$
(d) $y= (12 x^{3}+4)^{1/4}$

#### Work Step by Step

Given
$$y^{3} y^{\prime}-9 x^{2}=0$$
(a) Then
\begin{aligned} y^{3} \cdot \frac{d y}{d x} &=9 x^{2} \\ y^{3} d y &=9 x^{2} d x \end{aligned}
(b) Integrate both sides
\begin{aligned} \int y^{3} d y &=\int 9 x^{2} d x \\ \frac{y^{4}}{4} &=9 \cdot \frac{x^{3}}{3}+C \\ \frac{y^{4}}{4} &=3 x^{3}+C \end{aligned}
(c) Then $$ y= (12 x^{3}+C)^{1/4}$$
(d) At $x= 1$, $y=2$, then $C=4$, hence
$$ y= (12 x^{3}+4)^{1/4}$$