Why this variety is non-normal? According to Shafarevich, the "simplest possible example of a non-normal variety" is the curve $ X $ defined by $ y^{2} = x^{2}+x^{3}. $ 
Why is this variety non-normal? Furthermore, why is it the simplest possible non-normal variety?
Every nonsingular variety is normal, but I don't see how this helps me since $ X $ is singular.
 A: As a variation on the comment of reuns, let me explain why this curve isn't normal just using the definiton, without mentioning anything about Serre's criterion (which I consider to be more advanced).
The coordinate ring of your curve is $$R:= \frac{k[x,y]}{\left< y^2-x^2-x^3\right>}$$
Now the field of fractions of this ring contains the element $f=\frac{y}{x}$ which is integral over $R$, since it satisfies the equation
$$f^2-x-1=0.$$
On the other hand, you can check that $f$ itself does not belong to $R$. So $R$ is not integrally closed.
(Geometrically you can think of $f$ as giving the slope of the line joining the origin to a given point. Since the curve has two branches passing through the origin with distinct tangent lines, there is no way to extend $f$ to the origin.)
As for why this is the "simplest": well, it's a 1-dimensional variety defined by a single equation of degree 3. (If the degree were 1 or 2, such a variety would be nonsingular, hence normal.)  The only other reasonable candidate I can think of for "simplest" would be the cuspidal curve $y^2=x^3$.
A: A variety is normal iff it equals its normalization. In dimension $1$, normalization is a resolution of singularities. Thus normal also implies non-singular in the case of curves. You can see that by using Serre's criterion for normality.
The general rule is: ”Normal varieties are regular in codim $1.$“ That means for example, that surfaces do not contain singular curves, only singular points.
