For a function $f$, the definition of the Taylor Series of $f$ is: $\text{Taylor}(f)=\sum_{n \ge 0}\frac{f^{(n)}(a)}{n!}(x-a)^n, $ where $f^{(n)}$ indicates the $n$th derivative of $f$ with $f^{(0)}=f$.
The problem, then, reduces to the following: Is there an $n$th derivative of $f(x)=\frac{6}{x}$? (More aptly, since $6$ is a constant: Is there an $n$th derivative of $\frac{1}{x}$? If so, what is it?)
Let's do a table of values: $ \begin{align} \frac{d}{dx}\frac{1}{x}&=-\frac{1}{x^2}\\ \frac{d^2}{dx^2}\frac{1}{x}&=2\frac{1}{x^3}\\ \frac{d^3}{dx^3}\frac{1}{x}&=-6\frac{1}{x^4}\\ &\ldots \end{align} $ This pattern seems to suggest $\frac{d^n}{dx^n}\frac{1}{x}=(-1)^{n}n!\frac{1}{x^{n+1}}.$
Let's prove this via induction on $n$. The base case trivially true. Assume the induction hypothesis for $n=k$. $ \begin{align} \frac{d^{k+1}}{dx^{k+1}}\frac{1}{x}&=\frac{d}{dx}\frac{d^k}{dx^k}\frac{1}{x}\\ &=\frac{d}{dx}(-1)^kk!\frac{1}{x^{k+1}}\\ &=(-1)^kk!(k+1)(-1)\frac{1}{x^{k+2}}\\ &=(-1)^{k+1}(k+1)!\frac{1}{x^{k+2}}. \end{align} $
Thus, via induction, we see this is correct!
Can you take this from here? Apply the $n$th derivative and the definition of $\text{Taylor}(f)$.
Okay, so you have the following pieces of information: $a=-4$, $f^{(n)}(x)=(-1)^nn!\frac{1}{x^{n+1}}$, and the definition of the Taylor Series. You simply substitute these into the definition as follows:
$ \text{Taylor}(f)=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{a^{n+1}}}{n!}(x-a)^n=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{(-4)^{n+1}}}{n!}(x-(-4))^n . . . $
Try to go from here and simplify this. :)