How can I see mathematically that these two singularities are different? I came across these two curves while reading the wikipedia page on singularity theory:
$$
y^2=x^3+x^2
$$
and
$$
y^2=x^3
$$
The page says the cusp at $(0,0)$ can be seen to be qualitatively different but does not explain why they are different (other than visually). Is there a way to use the implicit function theorem? 
Both partials vanish at the origin of course, but I was wondering if I could gain intuition by something like speed of decay (the first curves partial in x decays faster, but it doesn't look as "nasty" so I am not sure if this even makes sense).
Edit: To clarify this last bit, the derivative taking $x$ to be endogenous of the two curves are
$$
y_1'=\frac{3x^2+2x}{2\sqrt{x^3+x^2}}\\
y_2'=\frac{3}{2}\sqrt{x}\\
$$
and while they are both bounded as you approach 0 from the right, the second has unbounded growth near zero. Does this mean anything?
 A: Both curves have a double point at the origin, but they're of different type: in $y^2=x^3+x^2$ the origin is an “ordinary” double point with distinct tangents, in $y^2=x^3$ the origin is a cusp, meaning there is a single tangent.
You can see it with a simple strategy: if we intersect the first curve with a line $y=mx$, we get the equation
$$
x^3+(1-m^2)x^2=0
$$
that intersects the curve at least twice, but there are three coincident roots for $m=\pm1$. These are the two tangents at the origin.
In the case of $y^2=x^3$ the equation is
$$
x^3-m^2x^2=0
$$
that has three coincident roots only for $m=0$.

A: An interesting note I found revisiting this problem:
For the first curve, which graphically has two tangents at $x=0$, when finding the derivative we get
$$
\frac{dy}{dx}=\frac{3x^2+2x}{2\sqrt{x^3+x^2}}=\frac{x(3x+2)}{2|x|\sqrt{x+1}}
$$
then the limit as $x\rightarrow 0$ depends on the sign of $x$
$$
\lim_{x\rightarrow 0^+}\frac{x(3x+2)}{2|x|\sqrt{x+1}}=1\\
\lim_{x\rightarrow 0^-}\frac{x(3x+2)}{2|x|\sqrt{x+1}}=-1
$$
And we get two tangents at 0. 
However for the cuspidal curve, we get
$$
\frac{dy}{dx}=\frac{3x^2}{2\sqrt{x^3}}=\frac{3}{2}\sqrt{x}
$$
Which is defined for $x$ positive. 
This gives me a sense of why the singularities should be different, I believe. Would appreciate feedback however.
