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.

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.
