Is the $f(x,y,z)$ is continuous and differentiable at $(0,0,0)$? 
Let
  $$f(x,y,z)=\begin{cases} 
\displaystyle\frac{xyz}{x^2+y^2+z^2}& \text{if $(x,y,z)\neq(0,0,0)$,}\\
0& \text{if $(x,y,z)=(0,0,0)$.}\end{cases}$$
  Is $f(x,y,z)$ continuous and differentiable at $(0,0,0)$?

In case of two variables I know that we can approach to the orig from different curves. in this case can we approach to the origin from two different curves at the same time? Any help is appreciated.
 A: By the AGM-inequality
$$|xyz|^{2/3}\leq \frac{x^2+y^2+z^2}{3}.$$
Hence, as $(x,y,z)\to (0,0,0)$,
$$|f(x,y,z)|=\frac{|xyz|}{x^2+y^2+z^2}\leq \frac{(x^2+y^2+z^2)^{1/2}}{3^{3/2}}\to 0$$
and $f$ is continuous at $(0,0,0)$ because $f(0,0,0)=0$.
Moreover $f$ is zero along the coordinate axes, therefore, in order to have the differentiability we should show that
$$\lim_{(x,y,z)\to (0,0,0)}\frac{f(x,y,z)}{(x^2+y^2+z^2)^{1/2}}=\lim_{(x,y,z)\to (0,0,0)}\frac{xyz}{(x^2+y^2+z^2)^{3/2}}=0.$$
What happens to this ratio along the line $t\to (t,t,t)$ as $t$ goes to $0^+$?
A: Firstly we might check whether or not f is continuos in (0,0,0) by calculating the limit as $(x,y,z)\to(0,0,0)$. If it is discontinuos then it can't be differentiable indeed continuity is a necessary condition since differentiability implies continuity.
And in this case for  $(x,y,z)\to(0,0,0)$ by spherical coordinates we have
$$\displaystyle\frac{xyz}{x^2+y^2+z^2}=\rho \sin^2 \phi\sin \theta\cos \theta=0$$
To check differentiability we need to check by the definition that
$$\lim_{(h,j,k)\rightarrow (0,0,0)} \frac{ f(h,j,k)-f(0,0,0)-\nabla f(0,0,0)\cdot (h,j,k)}{\| (h,j,k)\|}=0$$
and since $\nabla f(0,0,0)=0$ we need to show that
$$\lim_{(h,j,k)\rightarrow (0,0,0)} \frac{hjk}{(h^2+k^2+j^2)^\frac32}=0$$
but as pointed out for some trajectory it is not true.
A: Continuous, yes. Intuitively, the numerator vanishes to 3rd order as $(x,y,z) \to 0$, but the denominator only vanishes to 2nd order, so you should be able to show that $\lim_{(x,y,z) \to 0} f = 0$.
Differentiable, no. As you say, you can approach the origin along different paths. If you approach along any coordinate axis, so a path like $\gamma(t) = (t,0,0)$, you should get that $f(\gamma(t)) \equiv 0$. So if $f$ was differentiable, you would have $\nabla f = 0$. But if instead you take the path $\gamma(t) = (t,t,t)$, you'll get a nonzero derivative, so this can't be correct.
