I want to figure out why $\{(t^2,t^3)| t\in\mathbb{R}\}$ is not a regular submanifold of $\mathbb{R}^2$. There are some posts discussing this (like this post). But I do not quite understand them. So I want to directly apply the definition. Here is my proof. Is it correct?
We claim that for $p=(0,0)$ in the above set, there is no diffeomorphism $\varphi: U=(x^2+y^2<\epsilon^2)\to V$, $(x,y)\mapsto (u(x,y),v(x,y))$ s.t. for $x^2+y^2<\epsilon^2$, $u(x,y)=0\iff x=t^2$, $y=t^3$ for some $t$. $\varphi$ is a diffeomorphism $\implies$ $u$ is $C^{\infty}$, and $u_x$, $u_y$ can not be both zero at $p=(0,0)$. In any neighbourhood of $(0,0)$, $x=t^2$, $y=t^3$ does not give a function $y(x)\implies\ u_y=0$ (otherwise this contradicts the inverse function theorem). Hence $u_x\ne 0$, and in some neighbourhood of $p=(0,0)$, $x=t^2$, $y=t^3$ gives a $C^{\infty}$ function $x(y)$ since $u$ is $C^{\infty}$. But $x(y)=y^{\frac{2}{3}}$ is not $C^{\infty}$. This is absurd. Hence such $\varphi$ does not exist, and $\{(t^2,t^3)| t\in\mathbb{R}\}$ is not a regular submanifold of $\mathbb{R}^2$.