Question:
What real values of $x$ are not in the domain of

$f(x)=\frac{1}{|x^2+3x-4|+|x^2+9x+20|}$?

Answer:
$x$ is not in the domain of $f$ if the denominator is zero.  Since both absolute values are nonnegative, both must be zero for the denominator to be zero.  So

\begin{align*}
0=x^2+3x-4=(x+4)(x-1)&\Rightarrow x=-4\text{ or }x=1\\
0=x^2+9x+20=(x+4)(x+5)&\Rightarrow x=-4\text{ or }x=-5
\end{align*}

The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.