Differential equation solved as a separable equation and the solution y=constant I have a doubt in the definition of the solution of the differential equations in general. For this question we can use the following differential equation:
$$
y'(x)=3x^2y^2
$$
We can solve it as a separable equation, where:
$$
y'(x)=a(x)b(y)
$$
And in our case:
$$
a(x)=3x^3, b(y)=y^2
$$
Now, when I solve the differential equation I get:
$$
\int \frac{1}{y^2}dy=\int3x^2dx \Rightarrow -\frac{1}{y(x)}=x^3+C \Rightarrow y(x)=-\frac{1}{x^3+C}
$$
But if we go back to the original differential equation we can see that $ b(0)=0 $  and $y'(0)=0$.
So that is it right to say that the differential equation solutions are:
$$
y(x)=-\frac{1}{x^3+C};y(x)=0
$$
I have this doubt because on some books this step is important but for example, when I check my exercises on Wolfram Alpha the "constant solution" is ignored.
 A: $y(x)=0$ is a particular solution of $y'(x)=3 x^2 y^2$ , which general solution is $y(x)=\frac{1}{c_1-x^3}$
The particular solution $y(x)=0$ corresponds to $c_1=\infty$ and so, is included in the general form above.
I couldn't say if this interpretation is consistent with the WolframAlpha internal logic.  
IN ADDITION :
A few remarks about the solution $y(x)=0$ and the formulas expressing the general solution of the ODE.
Considering the solutions on the form $y(x)=\frac{1}{c_1-x^3}$ gives the false feeling that the particular solution $y(x)=0$ is on a different status than the other particular solutions because it is given by the formula for $c_1=\infty$, as it was a solution at limit. This is subject to discussion.
In fact, in a formula incuding an arbitrary constant, for example $c_1$, the constant can be remplaced by another constant on a different form. For example, instead of $c_1$, it could be $\frac{1}{c_2}$ and the formula becomes :
$$y(x)=\frac{c_2}{1-c_2x^3}$$
One can prove that the new formula is as well convenient by puting it into the ODE. Now, consider the case $c_2=0$. It gives the particular solution $y(x)=0$ without the need to invoque a limit. So, the solution $y(x)=0$ is one among all the particular solutions, without requiring any special status.
One can observe that the formula $y(x)=\frac{c_2}{1-c_2x^3}$ gives the particular solution $y(x)=-\frac{1}{x^3}$ for $c_2=\infty$ , again invoquing a limit. This particular solution was given for $c_1=0$ by the first formula $y(x)=\frac{1}{c_1-x^3}$, without limit.
Another form to express the general solution of the ODE can be derived for example by remplacing $c_1$ by $\frac{c_1}{c_2}$ leading to :
$$y(x)=\frac{c_2}{c_1-c_2x^3}$$
Again, one can prove that the new formula is as well convenient by puting it into the ODE. The particular solution $y(x)=-\frac{1}{x^3}$ corresponds to $c_1=0$. The particular solution $y(x)=0$ corresponds to $c_2=0$. Now, both particular solutions are obtained without limit.
I understand that one could prefer the form $y(x)=\frac{c_2}{c_1-c_2x^3}$ to the form $y(x)=\frac{1}{c_1-x^3}$ or to the form $y(x)=\frac{c_2}{1-c_2x^3}$ . On my own feeling, it doesn't matter, they are equivalent. 
A: One of possible sources of misconceptions lies in the erroneous application of separation of variables. The key point that you should apply it to Cauchy problems, not to the equation itself.
In general case, when we work with $$y'(x)=a(x)b(y),\quad y(x_0)=y_0.$$
You want that $b$ is Lipchitz continuous around $y_0$ and $a$ continuous around $x_0$ - hence, you can guarantee local existence and unicity of solutions.
Then again, if $b(y_0)=0$, then the solution $y(x)=y_0$ is a stationary solution and we are done.
If $b(y_0)\ne 0$, then $b(y)\ne 0$ in some neighbourhood of $y_0$ and we can divide everything by $b(y)$ to obtain
$$\frac{y'(x)}{b(y(x))}=a(x).$$
Now we integrate - but we can only consider the definite integral to find the solution:
$$\int_{y_0}^{y(x)}\frac{ds}{b(s)} = \int_{x_0}^{x}a(s)ds.$$
If $B(s)$ is the antiderivative of $1/b(s)$ and $A(s)$ - the antiderivative of $a(s)$,  then you can say that
$$B(y(x))=A(x)-A(x_0)+B(y_0).$$
Afterwards, you want to invert $B$ to obtain the value of $y(x)$.
In our case if $y_0=0$, then we obtain the stationary solution that falls to the first category. Otherwise, we obtain the integral
$$\int_{y_0}^{y(x)}\frac{ds}{s^2} = 3\int_{x_0}^{x}s^2ds$$ 
$$ \frac{1}{y_0}-\frac{1}{y(x)}  = x^3-x_0^3,$$
which results into 
$$y(x) = \frac{1}{1/y_0+x_0^3-x^3} = \frac{y_0}{1+y_0(x_0^3-x^3)}.$$
This form of the solution allows us to investigate the maximum domain of existence, time and rate of explosion of the solution, etc. Moreover, the continuous dependence of solution allows us to say - in some way - that for $y_0=0$ the solution is indeed zero. 
