Question: I just started working with partial fractions and hit a wall with splitting this one:

$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$

I get here:

$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$

Then on to:

$$ (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2 $$

I find that $E = 1$ by using $x = -2$. I am unsure how to proceed from here.

Answer: 
You already have 

\begin{align\*}
&\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}\\
&= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}\\ 
&= \frac{(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2}{(x+2)(x^2+x+1)^2}.
\end{align\*}

Multiplying both sides by $(x+2)(x^2+x+1)^2$ we have $$3x^2 + 2x + 1 = (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2.$$ The right hand side is a quartic equation. If you expand all the brackets, you can collect all the like terms to get an equation as follows: $$3x^2 + 2x + 1 = k\_4x^4 + k\_3x^3+k\_2x^2+k\_1x+k\_0$$ where $k\_i$ depends on $A, B, C, D,$ and $E$. Then you can compare the coefficients of $1, x, x^2, x^3,$ and $x^4$ to get five equations in five unknowns which will be enough to determine $A, B, C, D,$ and $E$.
You don't have any equations there.

Presumably, you start with
$$\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} 
= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}
$$
then multiply both sides
by $ (x + 2)(x^2 + x + 1)^2$
to get
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2
.$$

There are a variety of ways
to go from here.

As you did,
when you set
$x = -2$,
you get
$12-4+1=9E$
so $E = 1$.

The equation then becomes
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + (x^2 + x + 1)^2
.$$

You can do as
Michael Albanese
suggested to get equations for the
other coefficients.
This is probably the
best way.

You can look at the roots
of
$x^2 + x + 1=0$,
which are complex.

You can let $x=0$,
and this becomes
$1
=2B + 2D + 1
$,
or $B = -D$.

Looking at the coefficient of $x^4$,
you get
$Cx^4+x^4 = 0$
or $C = -1$.

I'll stop here.