Two different answers when computing dimension of Zariski cotangent space? Let $V = V(X^2-Y^3,Y^2 - Z^3)$. Let $P= (0,0,0)$ that is a point in $V$. I want to compute $\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2$ where $\mathfrak{m}_P$ is the maximal ideal in the local ring $$ \mathcal{O}_P(V) := k[V]_{(\overline{X},\overline{Y},\overline{Z})}.  $$
By Lemma 1.31 in Milne's notes here, it sufices to compute the dimension of $I/I^2$ where $I = (\overline{X},\overline{Y},\overline{Z})$ since $\mathfrak{m}_P = I_{(\overline{X},\overline{Y},\overline{Z})}.$ Now on one hand it is clear to me that $\dim_k I/I^2 = 3$. However suppose we consider the map
$$\begin{eqnarray*} \alpha: &k[X,Y,Z]& \to k[T]  \\ 
&(X,Y,Z)& \mapsto (T^9,T^6,T^4). \end{eqnarray*}$$
This induces an isomorphism $\overline{\alpha}$ between $k[V]$ and a subring of $k[T]$ since $$I(V(X^2-Y^3,Y^2 - Z^3))= (X^2-Y^3,Y^2 - Z^3).$$
The image of $I$ under $\overline{\alpha}$ is $(T^4,T^6,T^9) = (T^4)$ while $\overline{\alpha}(I^2) = (T^8)$. This means that
$$\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2 = \dim_k (T^4)/(T^8) = 4.$$


My question is: How can I get two different answers by trying to compute the dimension in two different ways?


 A: Well, one of the answers is obviously wrong: the tangent space can't have dimension greater than the ambient affine space. The homomorphism $\overline{\alpha} : k[V] \to k[t]$ you construct is also not an isomorphism. (What maps to $t$?)
A: Your variety $V$ is a curve whose only singularity is  the origin $O=(0,0,0)$.     
Its desingularization (=normalization) is the morphism $$a:\mathbb A^1_k\to V: t \mapsto (x,y,z)=(t^9,t^6, t^4)$$ whose associated comorphism is your $k$-algebra morphism $\alpha: k[V] \to k[T]$.    
The crucial point is that $a$ is a bijective morphism but not an isomorphism: the inverse rational map $$a^{-1}:V \to \mathbb A^1_k:(x,y,z )\mapsto t=\frac {x}{z^2}$$ is an isomorphism between the open subsets  $V\setminus \{O\}\subset V$ and $\mathbb A^1_k \setminus \{0\}\subset \mathbb A^1_k$, but is not regular at $O$.  
Hence   the $k$-algebra morphism $ \alpha$ is not an isomorphism (since $a$ is not an isomorphism), it does not  induce an isomorphism of $\mathfrak{m}_P/\mathfrak{m}_P^2$ with $(T^4)/(T^8)$ and thus, as already remarked by Zhen,  there is no contradiction.
