Estimates for solution of Laplace equation The function
    $\Phi(x):= \left\{\begin{array}{lr}
    - \frac{1}{2\pi} log(|x|), & \text{for } n=2\\
    \frac{1}{n*(n-2)\alpha(n)} \frac{1}{|x|^{n-2}}, & \text{for } n \ge  3
    \end{array}\right\} 
$ defines for $x \in \mathbb{R}^n, x \not=0$ is the fundamental solution of Laplace´s equation.
How i can observe that we hve the following esimates:$$|D\Phi(x)| \le \frac{C}{|x|^{n-1}},\quad\quad |D^2 \Phi(x)| \le \frac{C}{|x|^n} (x \not=0)$$
for some constant C>0.
I already calculate:     $(D \Phi(x))_i= \left\{\begin{array}{lr}
    - \frac{1}{2\pi} \frac{1}{|x|}\frac{x_i}{|x|}, & \text{for } n=2\\
    \frac{1}{n*(n-2)\alpha(n)} \frac{|x|^{-n+3}}{-n+3}\frac{x_i}{|x|}, & \text{for } n \ge  3
    \end{array}\right\} $ because $ \frac{\partial |x|}{\partial x_i}= \frac{x_i}{|x|}$ for $x \not=0$.
For n=2: $|D \Phi(x)|= \frac{1}{2\pi} \frac{|x|}{|x|^2}= \frac{1}{2\pi} \frac{1}{|x|} \le \frac{C}{|x|^{2-1}}$
But for n=3: $|D \Phi(x)|= \frac{1}{n(n-2) \alpha(n)*(3-n)} \frac{|x|}{|x|^{n-2}}=\frac{1}{n(n-2) \alpha(n)*(3-n)} \frac{1}{|x|^{n-3}}$

You did the calculation wrong for $n\geq 3$. We have
$$\nabla |x|^{2-n} = (2-n)|x|^{2-n-1}\frac{x}{|x|} = (2-n)\frac{x}{|x|^{n}}$$
