Maps between tangent space of $X=\mathbb{V}(x^3-y^2)$ and $Y=\mathbb{V}(x^4-y^3)$ It is an exercise, I am not sure if I am doing right.

Let $X=\mathbb{V}(x^3-y^2)$ and $Y=\mathbb{V}(x^4-y^3)$ in
  $\mathbb{A}^2$. 
1) Find the tangent spaces of $X$ and $Y$ at the origin, i.e. $T_0X$
  and $T_0Y$, and the natural bases for the dual vector space $(T_0X)^*$
  and $(T_0Y)^*$.
2) Consider the map $$ F:\mathbb{A}^1\rightarrow X, t\mapsto(t^2,t^3)
 $$
and map $$ G:\mathbb{A}^1\rightarrow Y, t\mapsto(t^3,t^4) $$
Show that $G$ factors through $F$ and write down a regular map
  $H:X\rightarrow Y$.
3) Using basis from 1) to find $$ dH_0:T_0X\rightarrow T_0Y $$

My approach is:
1)
$T_0X=T_0Y=\mathbb{A}^2$, and the bases of the dual spaces are both
$$
(x,y)\mapsto x, \text{and } (x,y)\mapsto y
$$
(Is it right? Looks wried...)
2)
Notice the map $\mathbb{A^1}\rightarrow X\rightarrow Y$ given by
 $t\mapsto(t^2,t^3)\mapsto(t^3,(t^2)^2)$, we have
$$
H:X\rightarrow Y, (x,y)\mapsto(y,x^2)
$$
3)
I am not sure about this part. The hint says using bases from 1), so I tried to define the dual map of $H$ by
$$
\begin{align*}
H^*:(T_0Y)^*&\rightarrow(T_0X)^*\\
[(x,y)\mapsto x]&\mapsto [(x,y)\mapsto y]\\
[(x,y)\mapsto y]&\mapsto [(x,y)\mapsto x^2]\\
\end{align*}
$$
But I don't know how to get $dH_0$ from this. Why can't I directly define $dH_0$ from the map of 2), i.e.
$$
dH_0:T_0X\rightarrow T_0Y, (x,y)\mapsto(y,x^2)
$$
Could anyone help?
 A: 1) For the tangent space I get the same, I'm not sure what is meant by the natural basis in this context, but what you wrote certainly is a basis, the one dual to your basis in the vector space. 
2) This seems to be alright; if you want to be complete, to get a map from $X$ to $Y$, you would have to say a word about why $t \rightarrow (t^2, t^3)$ is surjective onto $X$, but this is clear I think.
3) I'm also not sure about how this is meant, but what I can think of: My definition for the tangent space would be the dual of $m/m^2$ where $m$ is the maximal ideal in the local ring of X resp. Y at the origin, called $\mathcal{O}_{X,(0,0)}$ resp. $\mathcal{O}_{Y, (0,0)}$, which is in both cases $(x, y)/(x,y)^2$. The map $f$ you defined in 2.) induces a map $ \mathcal{O}_{Y,(0,0)} \rightarrow \mathcal{O}_{X, (0,0)}$ by composition with $f$, thus also induces a map on the dual of the tangent spaces. Using your basis from 1.), an element $ax + by \in (x,y)/(x,y)^2$ gets mapped to $ay + bx^2 = ay \in (x,y)/(x,y)^2$. Going back to the dual, i.e. the tangent space, this gives $(x, y) \rightarrow (y, 0)$ as the differential (btw. I think your answer cannot be correct, because the differential is a linear map of vector spaces).    
