I know there is alot of information at internet, saw a topic here related this concept.
But I like to make sure I understand.
A)
Regarding $f\left(z\right)=z\sin\left(\frac{1}{z}\right)$
We know it has singularity problem at $z=0$, so we can know if it is analytic at 0 or not by knowing it residue. By using taylor series, we will receive that $c_{-1}=0$ ( it has only even negative power). Does it mean I can say that $f(z)$ is entire in $\mathbb{C}$? since its residue is 0 and thus it has a antiderivative ( If I remember correctly ).
B)
Same with $g\left(z\right)=\frac{\sin\left(z\right)}{z}$, from its taylor series we can say pretty much that it is entire is $\mathbb{C}$ ( that I Know for sure, I proved, but just to make sure here ).
C)
Regarding $h\left(z\right)=\frac{z}{\sin\left(\frac{1}{z}\right)}\:$.
This function is a little problematic for me, according to studies, this function is not analytic in $0$ and at $\frac 1 {\pi k}$, for $k\in\mathbb{Z}$.
However, from its taylor series we will actually receive that $$\frac{z}{\frac{1}{z}-\frac{3!}{z^3}+O\left(z^5\right)}$$ and it is actually a number that has $\frac{p\left(z\right)}{q\left(z\right)}$ such that $deg\left[p\left(z\right)\right]>deg\left[q\left(z\right)\right]$.
So how is it possible it is not analytic at 0? or the other infinite points.
What am I missing here? I do not know why I am having hard time understanding this concept.
Let's quickly review your points.
A) $f(z) = z \sin(1/z)$ actually has an essential singularity at $z = 0.$ You can see this by using the power series expansion that defines $\sin z$ around $z = 0$ to obtain the Laurent series for $f$ around this point: $$f(z) = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n}}.$$ As you correctly saw, the residue of $f$ at $z = 0$ is $0,$ but since $f$ has infinitely many negative-power terms, it has an essential singularity there, so the residue is not that useful here.
B) $g(z) = \sin (z) / z$ has a removable singularity at $z = 0,$ since it has limit at that point, so you can actually extend $g$ by defining $g(0) := 1,$ and in this way it becomes an entire function.
C) $h(z) = z / \sin(1/z)$ cannot be analytic at $z = 0$ because this point is not an isolated singularity, i.e., you cannot find an open disk around $z = 0$ such that $f$ is analytic on the punctured disk. You can see this by noting that every point in the set $$\left\{ \frac{1}{k\pi} \, \middle\vert \, k \in \mathbb{Z} , \, k \neq 0 \right\}$$ is a simple pole of $\sin(1/z),$ and that $0$ is an accumulation point of this set. There's not much work to be done once the singularity is not isolated, as far as I know.