Is there a mistake in this solution by my teacher? I have this problem :

let $\alpha > 0$ and $f:\mathbb{R^2}\rightarrow \mathbb{R^2}$ such that $f(x,y)=(x^3-3x|y|^{\alpha}, 3|x|^{\alpha}y-y^3)$.
i) for which values of $\alpha$ is f differentiable at $(0,0)$
ii) for which values of $\alpha$ is f differentiable at $(1,0)$

So for i) the solution states that $x^3$ and $y^3$ are $C^{\infty}$ and $3x|y|^{\alpha}$ and $3y|x|^{\alpha}$ are $o(||x,y||)$ for any $\alpha >0$, therefore differentiable at $(0,0)$
Now for ii) the solution says that the partial derivative of $x|y|^{\alpha}$ doesn't exist at $(1,0)$ for $\forall \alpha \leq 1$, so for any $\alpha > 1$ $f$ is differentiable at (1,0). So here is where I doubt and I don't understand why by the same argument $f$ isn't differentiable at $(0,0)$?
Here's my reasoning :

let $g:=x|y|^{\alpha}$ we can rewrite $g$ as $g(x,y) = x[\sqrt{y^2}]^{\alpha}$. Then $lim_{x,y\rightarrow(0,0)} \frac{\partial g}{\partial y}(x,y)= lim_{x,y\rightarrow(0,0)}  \alpha x \frac{|y|^{\alpha}}{y}$ doesn't exist for $\alpha = \frac{1}{2}$ (we take $y = x^2$ and the limit diverges as $(x,x^2)\rightarrow (0,0)$

Doesn't this reasoning imply that $f$ is differentiable at $(0,0)$ iff $\alpha > 1$ ?

What you've shown is that the partial derivatives are continuous at $(0,0)$ if and only if $\alpha > 1$.
However, continuity of the partial derivatives provides only a sufficient test for differentiability (it's not necessary).  That is, from your argument we could conclude $f$ is differentiable at $(0,0)$ for $\alpha > 1$, but we cannot conclude anything for $0 < \alpha \leq 1$.  For that, we really need to use your teacher's method.

Note that your teacher's solution for part ii uses whether the partial derivative exists at the point $(1,0)$.  This is a necessary condition for differentiability at $(1,0)$ which is why they're able to conclude it's not differentiable at $(1,0)$ for $\alpha \leq 1$.  Arguably they haven't shown differentiability at $(1,0)$ for $\alpha >1$, so it's incomplete, but they may have felt that's the simpler case (following from continuity of the partials).
