<s>[INST] Derivatives in $\mathbb{R}^n$ 
Question: Let $~f : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$, $x \mapsto \frac{x}{x_1^2 + \dots + x_n^2}$. 
Show that $~f$ is differentiable at every point $x \in \mathbb{R}^n \setminus \{0\}$ and compute the Jacobian. Then
  provide evidence of whether it is possible to extend $f$ to a
  continuous function on $\mathbb{R}^n$.

What I have so far is that we can let $\mathbf{f} = (f_1, \dots, f_n)$, where
$$f_1 = \frac{x_1}{x_1^2 + \dots + x_n^2},~~f_2 = \frac{x_2}{x_1^2 + \dots + x_n^2},\dots,~~f_n = \frac{x_n}{x_1^2 + \dots + x_n^2}.$$
Since these are rational functions the partials exist, and from $(x_1, \dots, x_n) \neq (0, \dots, 0)$ we have that $x_1^2 + \dots + x_n^2 \neq0$. Meaning, all functions $\mathbf{f}$ are also continuous (not positive this exactly holds?). Hence, we have partials of form
$$\begin{align*}
\frac{\partial f_1}{\partial x_1} &= \frac{(x_1^2 + \dots +x_n^2) - 2x_1(x_1)}{(x_1^2 + \dots + x_n^2)^2}  = \frac{x_1^2 + \dots +x_n^2 - 2x_1^2}{(x_1^2 + \dots + x_n^2)^2} = \frac{x_2^2 + \dots +x_n^2 - x_1^2}{(x_1^2 + \dots + x_n^2)^2} \\
\frac{\partial f_1}{\partial x_2} &= \frac{-2x_2(x_1)}{(x_1^2 + \dots + x_n^2)^2} = \frac{-2x_1x_2}{(x_1^2 + \dots + x_n^2)^2} \\
& \vdots \\
\frac{\partial f_1}{\partial x_n} &= \frac{-2x_n(x_1)}{(x_1^2 + \dots + x_n^2)^2} = \frac{-2x_1x_n}{(x_1^2 + \dots + x_n^2)^2},
\end{align*}
$$
and so on for each $f_1, \dots, f_n$. 
Then we can construct the Jacobian such that
$$D_{\mathbf{f}}(\mathbf{x}) = 
\begin{pmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \dots & \frac{\partial f_1}{\partial x_n} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n} \\
\vdots & \vdots & \ddots  & \vdots \\
\frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \dots &\frac{\partial f_n}{\partial x_n}
\end{pmatrix}.$$
Is this correct up until this point? How do I extend this $f$ to a continuous function on $\mathbb{R}^{n}$ (if possible). It seems we already have shown for $\mathbb{R}^n$?
 [/INST] Continuous extension is not possible. For suppose $f(0)=\alpha$, then $f(x)\rightarrow\alpha$ as $x\rightarrow 0$. But $|f(x)|=\left|\dfrac{x}{|x|^{2}}\right|=\dfrac{1}{|x|}\rightarrow\infty$ as $x\rightarrow 0$.
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