Background: Let $X = V(I)$ be a closed subvariety of $Y = \mathbf{A}_k^n$ and let $x \in X(k)$ be a rational point, i.e. $k(x) = k$ (residue field). We can identify $E = k^n$ with $T_{Y,y}$. Let $f: X \to Y$ denote the closed immersion. Then we can identify $T_X,x$, the (Zariski) tangent space, with $$\left\{ (t_1, \dots , t_n) \in E \mid \sum _{1 \le i \le n} \frac{\partial P }{\partial T_i} (x) t_i = 0, \forall P \in I \right\}$$.
Now let's specialize to a particular example.
With the identification above, I can see easily that $\dim T_{X,x_0} = 2$ since the differential is always zero and therefore, the above set is equal to $k^2$ (dimension 2).
Now suppose $x \ne x_0$. Then the differential form is not zero so it seems that the above set should be equal to just the point $0 \in k^2$ which would have dimension 0, not 1. Can someone resolve this conflict?
Finally, I have computed, or at least attempted to compute, the dimension of the Zariski tangent space via this identification, rather than attempting to compute the dimension of $m/m^2$. Could one attempt to compute the dimension of $m/m^2$ directly? If so, how can we do it for this example?
$\{(t_1,t_2) \mid \frac{\partial P }{\partial T_1} (x) t_1 + \frac{\partial P }{\partial T_2} (x) t_2 = 0\}$ is the set of vectors in $E$ orthogonal to $\vec{v}_x= (\frac{\partial P }{\partial T_1} (x) , \frac{\partial P }{\partial T_2} (x) )$, which has codimension $1$ when $\vec{v}_x\neq 0$ and codimension $0$ when $\vec{v}_x=0$.
One of the easiest ways to compute $\mathfrak{m}/\mathfrak{m}^2$ directly is to identify $k[T,S]/(T^2-S^3)$ with the subring $k[X^2,X^3]$ of $k[X]$. If $\mathfrak{m}=(X-a)\neq (X)$, then $X\notin\mathfrak{m}$, so $k[X^2,X^3]_\mathfrak{m} = k[X]_{\mathfrak{m}}$.
Since $\mathfrak{m}/\mathfrak{m}^2$ only depends on the localization at $\mathfrak{m}$, this lets us compute in $k[X]$: $\mathfrak{m}/\mathfrak{m}^2 = (X-a)/(X-a)^2$ is easily seen to have dimension $1$ over $k$.