See my comment:
The differential equation is only valid on the following continuous intervals: 1) $-\infty < y < 0$, 2) $0 < y < 2$, and 3) $2 < y < \infty$. It might be valid on more but since $1 + 3x^2 = 0$ permits no real solutions, there's no need to worry about possible removeable discontinuities.
Find the "solution" first (which you did):
(3y^2 - 6y)dy =&\ (1 + 3x^2)dx \\
y^3 - 3y^2 =&\ x + x^3 + C
Plug in $y = 1$ and $x = 0$ to find:
1 - 3 = C \rightarrow C = -2
This gives the following:
y^3 - 3y^2 = x + x^3 - 2
(this is what you found)
Since $y(0) = 1$ we know that we must be on the interval $0 < y < 2$. Find the min and max on this interval: max of $f(y) = y^3 - 3y^2$ on the interval $0 < y < 2$. We already know that the critical points will be $y = 0$ and $y = 2$--the end points. So just input those two numbers. This tells us that the right side ranges from $y = 0$, $0^3 - 3\cdot0^3 = 0$ to $y = 2$, $2^3 - 3\cdot2^2 = 8 - 12 = -4$. So the $x$ values, $x + x^3 - 2$, range from $-4$ to $0$.
We need to find the continuous interval of $f(x) = x + x^3 - 2$ which is bounded by $-4 < f(x) < 0$. You can verify that there are no (real) extrema (meaning it's either monotonically increasing or decreasing) of the function $f(x) = x + x^3 - 2$ by showing that $f'(x) = 1 + 3x^2 = 0$ has no real solutions. This means that the interval is simply finding the two following solutions:
x + x^3 - 2 = -4 \\
x + x^3 - 2 = 0
Clearly $x = 1$ solves $x + x^3 - 2 = 0$. Also it's clear that $x = -1$ solves $x + x^3 - 2 = -4$. This means the $x$ interval is $-1 < x < 1$ or $|x| <1$. There is no "algebraic" way (that I'm aware of) to solve those two equations other than Rational Zeros Theorem (which means your particular problem is very special--i.e. it admits rational solutions, which will not generally be the case). Otherwise, it requires numerical methods.