The space $x^3-y^2=0$ Consider $\{(x,y)\in\mathbf{R}^2 \ | \ x^3-y^2=0\}$ as a subspace of $\mathbf{R}^2$. Intuitvely I understand that this is not supposed to be a differentiable manifold because it has a cusp at $0$. But how to prove this formally, is there a clean way to do this? Is this even a topological manifold? Thanks in advance.
a) Yes your curve $C=\{(x,y)\in\mathbb R^2 \ | \ x^3-y^2=0\}$ is a topological manifold because it is homeomorphic to $\mathbb R$ by the homeomorphism $\mathbb R\to C :t\mapsto (t^2,t^3) $, whose inverse is $C\to \mathbb R: (x,y)\mapsto \sqrt [3] y$.  

b) No, $C$ is not a differential submanifold of $\mathbb R^2$:   

Consider any  differentiable map $\gamma:\mathbb R\to C:t\mapsto (u(t),v(t))$ with $\gamma(0)=(0,0)$ and of course $$u^3-v^2=0 \;(\bigstar )$$  Since $u$ has a minimum at $t=0$ (all points of $C$ satisfy $x\geq0$ !) we have $u'(0)=0$.
On the other hand since $u(0)=0, v(0)=0$ we can write $u=tU, v=tV$  with $U,V$ differentiable and satisfying $u'(0)=U(0), v'(0)=V(0)$.
Substituting into $(\bigstar)$ and cancelling $t^2$ we  see that $V^2-tU^3=0$, so that $V^2(0)=0$, hence  $V(0)=v'(0)=0$ and thus finally $\gamma'(0)=(u'(0),v'(0))=(0,0)$.
This proves that $C$ is not a manifold because if it were we could find such a differentiable  $\gamma$ with $\gamma(0)=(0,0)$ but $\gamma'(0)\neq (0,0)$.
The map $(x,y)\mapsto (x^3-y^2,y)$ is a homeomorphism of $\mathbb{R}^2$ with inverse $(a,b)\mapsto ( (a+b^2)^{\frac{1}{3}}, b)$ and takes $C$ to the vertical $a=0$, so $C$ is a topological manifold.
We'll give a proof that $C$ does not have a regular parametrization around $(0,0)$ by a $C^1$ function ($(0,0)$ is not a regular point of $C$). First notice that if $O$ is a regular point on a curve then for $P$, $Q$ near $O$ we have $|P,Q| \simeq \text{length}(P,Q)$, the length of the curve from $P$ to $Q$, that is 
$$\frac{PQ}{\text{length}PQ}\to 1$$
as $P$, $Q$ approach $O$ on the curve.
. Now take two points $P_1$, $P_2$ on $C$, close to $O=(0,0)$ and symmetric. We have 


$$\frac{P_1P_2}{\text{length}(P_1,P_2)}=\frac{P_1P_2}{\text{length}(P_1,O)+\text{length}(O,P_2) }\le \frac{P_1P_2}{P_1O+OP_2}\to 0 $$
as $P_1$, $P_2 \to O = (0,0)$ (look at the cusp). Therefore $O$ is not a regular point of $C$.

In fact, for regular points $O$ and $P_1$, $P_2$ on both sides of $O$ and close to $O$, the points $P_1$, $O$, $P_2$ are almost in line, that is $\frac{P_1P_2}{P_1 O + OP_2} \to 1$, and that is  not true for cusps. 

Or, for regular points $O$ and $P_1$, $P_2$ on both sides of $O$ and close to $O$, the directions of the vectors $P_2 0, OP_1$ are almost the same. For this curve they become almost opposite.