The function is:$$f(x,y,z) = \begin{cases} x^2y^2z^2\sin(\frac{1}{xyz}), & \text{if $xyz\ne0$} \\ 0, & \text{if $xyz = 0$} \end{cases}$$
I need to say where is it differentiable, find the differential, and find at which points it is differentiable continuously.
At points where $xyz \ne 0 $, $f$ is a composition of two differentiable functions, thus it is differentiable.
Also I found that at each other point the partial derivatives equal zero, since for example: $$\lim\limits_{h \to 0} \frac{f(h,y,z)-f(0,y,z)}{h} = \lim\limits_{h \to 0} \frac{h^2y^2z^2\sin(\frac{1}{hyz})-0}{h} = \lim\limits_{h \to 0} hy^2z^2\sin(\frac{1}{hyz})=0$$
Now I know that the function is differentiable at those points if $$f(\mathbf{x}) - f(\mathbf{a}) - D(\mathbf{x}-\mathbf{a}) = o(\mathbf{x}-\mathbf{a})$$
where $D$ is the gradient which equals $(0,0,0)$ and $\mathbf{a}$ is a point where $xyz=0$, which means I need to show that: $$x^2y^2z^2\sin(\frac{1}{xyz}) = o(\mathbf{x}-\mathbf{a})$$ However that doesn't work for me. Does it mean that it is not differentiable at those points?
Moreover, by looking at the partial derivatives I came to the conclusion that $f$ is continuously differentiable at every point at which $xyz=0$, since the partial derivative, for example, contains the expression: $yz\cos(\frac{1}{xyz})$
Am I correct here?
The partial derivative doesn't just contain $yz\cos(1/xyz)$; it is $2xy^2z^2\sin(1/xyz) - yz\cos(1/xyz)$, where $xyz\ne0$. If the limit of this, as $xyz\to0$, is $0$ (the partial derivative where $xyz=0$), then $f$ is continuously differentiable where $xyz=0$; otherwise not. Actually, you need to do separately the points where one variable is $0$ (the coordinate planes), those where two variables are $0$ (the coordinate axes), and the point where all three are $0$ (the origin). By symmetry between the variables, it's continuously differentiable on the coordinate planes if the limit is $0$ as $x\to0$ with $y,z\ne0$ and as $y\to0$ with $x,z\ne0$; it's continuously differentiable on the coordinate axes if the limit is $0$ as $x,y\to0$ with $z\ne0$ and as $y,z\to0$ with $x\ne0$; and it's continously differentiable at the origin if the limit is $0$ as $x,y,z\to0$.
Wherever $f$ is continuously differentiable, it's differentiable, without having to check anything about $o(\mathbf{x}-\mathbf{a})$. For the places that you still need to check, by symmetry, you can assume that $x=0$ but $y,z\ne0$ at $\mathbf{a}$ (for $\mathbf{a}$ on a coordinate plane), $x,y=0$ but $z\ne0$ at $\mathbf{a}$ (for $\mathbf{a}$ on a coordinate axis), or $x,y,z=0$ at $\mathbf{a}$ (for $\mathbf{a}$ the origin). (But you won't actually have to all of these after the previous result.) So for example, if $x=0$ but $y,z\ne0$ at $\mathbf{a}$, you are looking at $\mathbf{a} = (0,y_1,z_1)$, so $\mathbf{x}=(x,y,z)$ with $x\to0$, $y\to y_1$, and $z\to z_1$, so $\lVert\mathbf{x}-\mathbf{a}\rVert = \sqrt{x^2+(y-y_1)^2+(z-z_1)^2}$, so you want to show that $$x^2y^2z^2\sin(1/xyz)/\sqrt{x^2+(y-y_1)^2+(z-z_1)^2} \to 0$$ as $(x,y,z)\to(0,y_1,z_1)$. Do you see how to do this?