Where is $f(x,y,z)=\begin{cases} 0 & \text{$xyx=0$} \\ x^2y^2z^2\sin \big (\frac{1}{xyz} \big) & \text{$xyz \ne 0$} \end{cases}$ continuously differentiable?
What I got so far is that $f$ can be written as a composition, $f=h\circ g$ where $g(x,y,z)=xyz$ and $h(t)=t^2\sin \big(\frac{1}{t} \big)$. By the chain rule we have that: $\nabla f(x,y,z) = h'(g(x,y,z))\cdot \nabla g(x,y,z) = \big (2xyz\cdot \sin \big( \frac{1}{xyz} \big) - \cos \big( \frac{1}{xyz} \big)\big)\cdot (yz,xz,xy)$.
Is that even correct? What should the gradient be at the origin or in other points where $xyz=0$? I'm really not sure how this fits into the mix. Also, is the function continuously differentiable in these points?
Yes, what you have is correct.
Now you need to consider the case, where
$$x = 0 \quad \lor \quad y = 0 \quad \lor \quad z = 0,$$
and then consider the function $$f = h \circ g.$$
After finding $Df$ piecewisely, you need to check whether that piecewise function, $Df$, is continuous or not.