Is $(x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}}$ Differentiable in $(0,0,0)$? $$f(x,y,z)=\begin{cases} (x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}} \qquad (x,y,z) \ne (0,0,0) \\ \\ 0 \qquad (x,y,z)=(0,0,0) \end{cases} $$ 
At first, I study the continuity in the origin. 
I apply the concept of sequential continuity: $x_n \rightarrow x_0 \Rightarrow f(x_n) \rightarrow f(x_0)$ 

So, $x_n=\frac{1}{n}  $
$$x_n \rightarrow 0 \Rightarrow f(x_n,x_n,x_n) \rightarrow f(0,0,0)=0$$
$$\lim_{n\rightarrow +\infty} \frac{3}{n^2} \ \sin \frac{1}{\frac{\sqrt{3}}{n}}=0 $$
How can I continue the study of differentiation?
Thanks!
 A: You can't study the continuity of $f$ at the origin by choosing one sequence $\mathbf{p_n} = (x_n,y_n,z_n)$ that approaches $\mathbf{0}$ in a very specific way ($x_n = y_n = z_n = \frac{1}{n}$) and checking that $f(\mathbf{p_n}) \rightarrow f(\mathbf{0})$. You need to check that this works for all sequences converging to $(0,0,0)$ or, equivalently, check the $\varepsilon-\delta$ definition. In your case, given $\varepsilon > 0$, if $|| \mathbf{p} - \mathbf{0} || = ||\mathbf{p}|| = ||(x,y,z)|| < \sqrt{\varepsilon}$ then
$$ |f(\mathbf{p}) - f(\mathbf{0})| \leq x^2 + y^2 + z^2 = ||(x,y,z)||^2 < \varepsilon $$
so indeed $f$ is continuous at $\mathbf{0}$. To check differentiability, check first whether the partial derivatives exist. For example,
$$ \frac{\partial f}{\partial x}(\mathbf{0}) = \lim_{h \to 0} \frac{f(h,0,0) - f(0,0,0)}{h} = \lim_{h \to 0} \frac{h^2 \sin \left( \frac{1}{h} \right)}{h} = \lim_{h \to 0} h \sin \left( \frac{1}{h} \right) = 0. $$
Similarly, the partial derivatives $\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ exist and vanish. This means that if $f$ is differentiable,  we must have $df|_{\mathbf{0}} = \mathbf{0}$. Let us check that this is indeed the case (here, $\mathbf{p} = (x,y,z)$):
$$ \lim_{\mathbf{p} \to \mathbf{0}} \frac{f(\mathbf{p}) - f(\mathbf{0}) - 0 \cdot x - 0 \cdot y - 0 \cdot z}{||\mathbf{p}||} = \lim_{\mathbf{p} \to 0} \sqrt{x^2 + y^2 + z^2} \sin \left( \frac{1}{\sqrt{x^2 + y^2 + z^2}} \right) = 0 $$
and so $f$ is differentiable at $\mathbf{0} = 0$ with vanishing differential / gradient.
A: It is much easier to analyze in spherical coordinates:
$$r^2 = x^2 + y^2 + z^2$$
$$\theta = \operatorname{atan2}(y,x)$$
$$\phi = \arccos\left({\frac z r}\right)$$
$$f(r, \theta, \phi) = \begin{cases} r^2 \sin \left({\dfrac 1 r}\right); \ r \ne 0  \\ 0; \qquad\qquad \  r = 0\end{cases}$$
Now you can use Calc I methods to determine continuity.
Edit: Amended according to levap's excellent comments, you need to examine the definition of differentiability on $f$ at $\mathbf 0$, since the spherical coordinate system fails at the pole. 
You need to check:
$$\displaystyle \lim_{\mathbf {r \to 0}}  {\frac{\vert f(\mathbf r) - f(\mathbf 0) - \mathbf {0 \cdot r}  \vert} {\vert \mathbf r \vert}} = \lim_{r \to 0} \left \vert {r \sin \left({\frac 1 r}\right)}\right\vert$$
and show it equals zero, which is the same as levap's answer.
