Proving differentiability (rigorously). 
How do I prove them to be differentiable in a "rigorous" way?
Or do i just simply compute their differentials?
 A: (i) Write $f=(f_1,f_2,f_3)$ and note it is differentiable if and only if each $f_j$ is differentiable. Now each $f_j$ is actually $C^\infty$ by composition/product/difference of standard $C^\infty$ functions. To express the derivative of $f$, compute the Jacobian.
(ii) This one is more tricky. It is $C^\infty$ on $\mathbb{R}^2\setminus \{(0,0)\}$ as a composition/product/sum/quotient of $C^\infty$ functions (where the denominator is nowhere zero). In this case, again, compute the Jacobian to express the derivative. You get
$$
df_{(x,y)}=\left(\frac{\partial f}{\partial x} \;\;\frac{\partial f}{\partial y}\right)=\left(\frac{2x^4+3x^2y^2}{(x^2+y^2)^{3/2}}\;\;\; \frac{-x^3y}{(x^2+y^2)^{3/2}}\right).
$$
At $(0,0)$, the argument is different. We will show that the derivative at this point is $0$. By definition, this amounts to showing that
$$
f((x,y))=f((0,0))+0\cdot(x,y)+\sqrt{x^2+y^2}\epsilon((x,y))
$$
where $\epsilon$ is a function which tends to $0$ as $\sqrt{x^2+y^2}$ goes to $0$.
So we compute
$$
\epsilon((x,y))=\frac{f((x,y))-f((0,0)-0\cdot(x,y)}{\sqrt{x^2+y^2}}=\frac{x^3}{x^2+y^2}.
$$
Now
$$
|\epsilon((x,y))|=\frac{|x|^3}{x^2+y^2}\leq \frac{|x|(x^2+y^2)}{x^2+y^2}=|x|\leq \sqrt{x^2+y^2}.
$$
So $\epsilon$ tends to $0$ as desired and $f$ is differentiable at $(0,0)$ with
$$
df_{(0,0)}=0.
$$
A: I think this depends on what type of reasoning the math course you're taking is trying to emphasize. For the first function, I would hope that you are allowed to take for granted the differentiability of $\sin$ and polynomials, in which case you just directly calculate according to the techniques you have learned.
For the second function, you can of course calculate the derivative for $(x,y) \neq (0,0)$. To show that it is also differentiable at the origin, you can use a $\delta,\epsilon$ type argument. Feel free to respond if you would like more elaboration.
