Given $\alpha>0$ define $F_\alpha:\mathbb R^n\to\mathbb R^n$ as $$F_\alpha(x)=x||x||^\alpha$$
I must prove that $F_\alpha\in\mathcal C^1(\mathbb R^n)$.
With some calculation I found that $$\frac{\partial(F_\alpha)_j}{\partial x_i}=\alpha x_jx_i||x||^{\alpha-2}=\alpha x_jx_i\frac{||x||^\alpha}{||x||^2}$$
Based on this result I would conclude that either $\alpha$ must be greater than $2$ (and not greater than $0$ as the exercise says) or $F_\alpha\in\mathcal C(\mathbb R^n\setminus\{0\})$.
What am I doing wrong?
Let us drop the subscript $\alpha$. It holds $$ \partial_i F_j (x) = \begin{cases} \delta_{ij} |x|^{\alpha} + \alpha x_i x_j |x|^{\alpha -2}, & \text{if}\ x\neq 0,\\ 0, &\text{if}\ x = 0. \end{cases} $$ (The partial derivatives at $x=0$ must be computed using the definition.)
It is easily seen that these partial derivatives are continuous. Namely, they are clearly continuous at points $x\neq 0$, whereas at $x=0$ it is sufficient to prove that $$ \lim_{x\to 0} \partial_i F_j (x) = 0. $$ But this condition is easily verified (for $\alpha > 0$) since $$ |\partial_i F_j(x)| \leq |x|^\alpha + \alpha |x|^2 |x|^{\alpha -2} = (1+\alpha) |x|^\alpha. $$