Show that the tangent surface of the twisted cubic curve $V(y-x^2,z-x^3)$ is not isomorphic to $\mathbb{R}^2$. This is an exercise in Ideals, Varieties and Algorithms by Cox et al.

Show that the tangent surface V of the twisted cubic curve $V(y-x^2,z-x^3)$, given by $$x=t+u\\y=t^2+2tu\\z=t^3+3t^2u$$ is not isomorphic to $\mathbb{R}^2$.

Following their hints, I found that $V$ is singular at all points on the twisted cubic curve. The singular points are where $f(x,y,z)=0$ and $\nabla f(x,y,z)=0$.
From here I can see $V$ is not isomorphic to $\mathbb{R}^2$ since $V$ is not smooth whereas $\mathbb{R}^2$ is. 
To prove that, the hint says that consider a polynomial map $\alpha: \mathbb{R}^2\rightarrow V$ such that $\alpha(a,b)$ is on the twisted cubic curve. Then the derivative matrix of $\alpha$ must have rank strictly less than $2$ at $(a,b)$. 
My question: With this mapping, we should have the surface defined by 
$$f(\alpha(u,v))=0$$
So $\nabla f \cdot \alpha'$ is the gradient of the surface on $u,v$. Shouldn't this be zero? And $\nabla f$ is zero at $(a,b)$ since it is singular on the twisted cubic. How does that tell the rank of the derivative matrix?
Thanks for any help!
 A: I've finally figured out a solution to this problem. But I still have a little doubt in my solution. I will be grateful if someone could explain it. 
Let $\alpha$ be the defined map. It is a parametrization of the surface $V$. So the derivative vectors $\vec{\alpha}_u$ and $\vec{\alpha}_v$ represents its tangent directions, and $\vec{\alpha}_u\times \vec{\alpha}_v$ is the normal direction, which is parallel to the gradient of $f$. 
At the singular points, where $(u,v)=(a,b)$, the gradient is zero. Hence the cross product of $\vec{\alpha}_u$ and $\vec{\alpha}_v$ is zero. This implies that the two tangential vectors are parallel. So the derivative matrix of $\alpha$ has rank less than $2$.
Now suppose $\beta: V\rightarrow \mathbb{R}^2$ is an inverse polynomial map. Then $\beta\circ \alpha$ is the identity map on $\mathbb{R}^2$. The derivative matrix of this map must be the $2\times 2$ identity, which is a contradiction. 
I believe this proof is complete. My doubt (a side thought) is, $\alpha\circ \beta$ is the identity map on $V$. Why does its derivative matrix not need to be identity? Even when $\alpha$ is not rank-deficient, it can only have rank $2$. Is it because $V$ is a 2D surface? I am not sure in which subject this is covered, maybe differential geometry?   
