I have to find the Taylor series of $\cos^3\!x$ around $0$. In this case, I have to find the Maclaurin series, which is: $$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+ \ldots +\frac{f^n(0)}{n!}x^n+ \ldots= \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$$
As I know I have to calculate the first, the second, the third deriavtive of the initial function, but how does it work in this case?
Assuming you know the series for $\cos x$, you could also use (power-reduction formula): $$\cos^3\!x = \tfrac{3}{4}\cos (x) +\tfrac{1}{4}\cos (3x)$$ where the series for $\cos(3x)$ simply follows from $\cos x$'s series after substitution $x \to 3x$:
$$\cos x = \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \quad\implies\quad \cos 3x = \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\left(3x\right)^{2n} = \sum\limits_{n=0}^{\infty} \frac{(-1)^n9^n}{(2n)!}x^{2n}$$ And then rearranging a bit: $$\begin{align} \cos^3\!x & = \frac{3}{4}\cos (x) +\frac{1}{4}\cos (3x) \\[6pt] & = \frac{3}{4}\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} +\frac{1}{4}\sum\limits_{n=0}^{\infty} \frac{(-1)^n9^n}{(2n)!}x^{2n} \tag{$*$} \\[6pt] & = \sum\limits_{n=0}^{\infty} \left( \frac{3}{4} \frac{(-1)^n}{(2n)!}x^{2n} +\frac{1}{4}\frac{(-1)^n9^n}{(2n)!}x^{2n} \right) \\[6pt] & = \sum\limits_{n=0}^{\infty} \left( \frac{1}{4}\frac{\left(3+9^n \right)(-1)^n}{(2n)!} \right) x^{2n} \end{align}$$ Since the series was obtained as a sum $(*)$ of two absolutely convergent power series with radius of convergence $+\infty$, the resulting series also has radius of convergence $+\infty$.