Answer
The solution of initial value problem is given by
$$
y^{3}-3y^{2}- x- x^{3}+2=0
$$
the interval in which the solution is valid is $| x | \lt 1$.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{(1+3x^{2})}{(3y^{2}-6y)} ,\quad \quad y(0)=1
$$
the differential equation can be written as
$$
(3y^{2}-6y)dy=(1+3x^{2})dx,\quad y(0)=1 \quad (*)
$$
integrating the left side with respect to $y$ and the right side with respect to $x$ gives
$$
y^{3}-3y^{2}=x+x^{3}+c \quad \quad (**)
$$
where c is an arbitrary constant.
To determine the solution satisfying the prescribed initial condition, we substitute $x=0$ and $y=1$ in eq. (**)
Hence the solution of initial value problem is given by
$$
y^{3}-3y^{2}- x- x^{3}+2=0 \quad \quad (**)
$$
The interval of validity of this solution extends
on either side of the initial point as long as the function remains differentiable, as we see in figure 1 which is attached.
We see that the interval ends when we reach points where the tangent line is vertical. It follows from the differential equation (*) that these are points where
$3y^{2}-6y =3y(y-2)=0$
i.e. when $y=0 $ or $y=2 $
and from equation (**) the corresponding values of $x$ are respectively
$x=1 $ or $x=-1 $ .
So the interval in which the solution is valid is $| x | \lt 1$.
