You are meant to just simply multiply the two series (in an informal manner) and keep the terms that have a power less than or equal to $6$.
It should be evident that you can consider the product $ \tag{1} \color{maroon}{ \Bigl( 1-x+ \frac{x^2}{2!}- \frac{x^3}{3!}+ \frac{x^4}{4!}- \frac{x^5}{5!}+ \frac{x^6}{6!}\Bigr)}\color{darkgreen}{ \Bigl( 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}\Bigr)}; $ because, for example, if you included the term $x^7\over7!$ of the series for $e^x$, then after doing the multiplication, that term would introduce powers of $x$ that are larger than 6, and you don't care about those.
So, let's do the multiplication. One way would be just to distribute the $\color{maroon}{\text{left}}$ factor of (1) across the $\color{darkgreen}{right}$, then keep distributing until no multiplications remain, and finally collect like terms.
But, a better method is available. You know you'll wind up with an expression of the form $\tag{2} a_0+a_1x^1+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6. $ We just need to figure out what the $a_i$ are.
Now, to obtain the product (1), you can do the following: select one term from the $\color{maroon}{\text{left}}$ factor and one term from the $\color{darkgreen}{right}$ factor and then take their product. Do this for every such choice of selections and add them all up.
So what we will do is, fix a value of $i$, and figure out the coefficient $a_i$ for the $x^i$ term appearing in the product (2).
So, how do you get $a_0$?
Well, there is only one way to obtain a constant by selecting terms from (1), and that is to select the "1" in both factors.
So $a_0=1$.
How do you get $a_1$?
Well, there is only one way to obtain a $x$ term by selecting terms from (1), and that is to select the "$\color{maroon}{-x}$" term from the $\color{maroon}{\text{left}}$ factor and the $\color{darkgreen}{1}$ from the $\color{darkgreen}{right}$.
So $a_1=1$.
How do you get $a_2$?
Here it's more interesting. There are exactly two ways:
$\ \ \ \ \bullet$select $\color{maroon}{1}$ from the $\color{maroon}{\text{left}}$ factor and $\color{darkgreen}{-{x^2\over 2!}}$ from the $\color{darkgreen}{right}$
$\ \ \ \ \bullet$select $\color{\maroon}{x^2\over 2!}$ from the $\color{maroon}{\text{left}}$ factor and $\color{darkgreen}{1}$ from the $\color{darkgreen}{right}$
Multiplying and adding these together will give $ 1\cdot{-x^2\over2!}+{x^2\over2!}\cdot1=0 $
So $a_2=0$.
Continuing in this manner:
$ \eqalign{ a_3&= \underbrace{-1\over2!}_{\color{maroon}{-x}\cdot \color{darkgreen}{x^2\over2!} } + \underbrace{-1\over3!}_{ \color{maroon}{-x^3\over3!}\cdot\color{darkgreen}1 } \cr a_4&= \underbrace{1\over4!}_{\color{maroon}1\cdot\color{darkgreen}{x^4\over4!}} -\underbrace{1\over2!2!}_{\color{maroon}{x^2\over2!}\cdot\color{darkgreen}{-x^2\over2!}} +\underbrace{1\over4!}_{\color{maroon}{x^4\over4!}\cdot\color{darkgreen}1} \cr a_5&= \underbrace{-1\over 4!}_{\color{maroon}{-x}\cdot\color{darkgreen}{x^4\over4!}}+ \underbrace{1\over3!2!}_{\color{maroon}{-x^3\over3!}\cdot\color{darkgreen}{-x^2\over2!}}+ \underbrace{-1\over5!}_{\color{maroon}{-x^5\over5!}\cdot\color{darkgreen}1} \cr a_6&= \underbrace{-1\over 6!}_{\color{maroon}1\cdot\color{darkgreen}{-1\over6!}}+ \underbrace{1\over2!4!}_{\color{maroon}{x^2\over2!}\cdot\color{darkgreen}{x^4\over4!}}- \underbrace{1\over4!2!}_{\color{maroon}{x^4\over4!}\cdot\color{darkgreen}{-x^2\over2!}} + \underbrace{1\over6!}_{ \color{maroon}{x^6\over6!}\cdot\color{darkgreen}1} \cr } $
Now you know the values of all the $a_i$ and you can form the required polynomial.
Note that, as Andre points out in the comments, your series for $e^{-x}$ is incorrect. The signs should be alternating: $e^{-x}=1-x+{x^2\over2!}-{x^3\over3!}+\cdots$.
I used the incorrect form in my answer before this edit; but have corrected it...