For $g(x) = \int_0^\infty (x+y)^{-1} f(y) \, dy$ show $|g'(x)| \le c_p \frac{1}{|x|^{1+1/p}} \lVert f \rVert_{L^p}$ 
Q: For $x > 0$ and $f \in L^P(0,\infty), 1 \le p < \infty$,
  \begin{align*}
    g(x) &= \int_0^\infty (x+y)^{-1} f(y) \, dy \\
\end{align*}
  Show that $g(x)$ is continuous and in fact differentiable with
  \begin{align*}
    |g'(x)| &\le c_p \frac{1}{|x|^{1+1/p}} \lVert f \rVert_{L^p} \\
\end{align*}

For $x \in (0,\infty)$:
\begin{align*}
    g'(x) &= \lim\limits_{h \to 0} h^{-1} (g(x + h) - g(x)) \\
    &= \lim\limits_{h \to 0} h^{-1} \int_0^\infty \left[(x+h+y)^{-1} - (x+y)^{-1}\right] f(y) \, dy \\
    &= \lim\limits_{h \to 0} - \int_0^\infty [(x+y)(x+h+y)]^{-1} f(y) \, dy \\
    &= - \int_0^\infty (x+y)^{-2} f(y) \, dy \\
\end{align*}
\begin{align*}
    |g'(x)| &\le \int_0^\infty |x+y|^{-2} |f(y)| \, dy \\
    &\le |x|^{-2} \lVert f \rVert_{L^1} \\
\end{align*}
It seems like I'm most of the way there. This looks like I might use Holder's inequality:
\begin{align*}
    \lVert fg \rVert_1 &\le \lVert f \rVert_p \lVert g \rVert_{\frac{p}{p-1}} \\
\end{align*}
but I don't see how.

You didn't properly justify bringing the limit inside the integral.
Once you have done so and obtain $|g'(x)| \le \displaystyle \int_0^\infty |x+y|^{-2} |f(y)| \, dy$ your thought to use Holder is correct. Note that
$$ \int_0^\infty |x+y|^{-2} |f(y)| \, dy \le \left( \int_0^\infty |x+y|^{-2p'} \, dy \right)^{1/p'} \|f\|_p.$$
Here
$$\int_0^\infty |x+y|^{-2p'} \, dy = \int_x^\infty |z|^{-2p'} \, dz = \frac{x^{1-2p'}}{1-2p'}$$
so that
$$
\left( \int_0^\infty |x+y|^{-2p'} \, dy \right)^{1/p'} = c_p x^{1/p' - 2} = c_p x^{-1-1/p}$$
