This isn't definitive, since the answer is always undefined, but let's be cutesy. Let's let $u' = a*c$ and $v = -a*c$, where $a < 1$
Then,
$$\begin{align}
u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\
&= \lim_{a\rightarrow 1}\frac{ac-ac}{1- \frac{a^{2}c^{2}}{c^2}}\\
&= \lim_{a\rightarrow 1}\frac{c-c}{-2a}\\
&= 0
\end{align}$$
Now, the reason why this isn't definitive is that you can take different limits, if you want. Say, let $u' = a^{2}c$ and $v = -ac$
Then,
$$\begin{align}
u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\
&= \lim_{a\rightarrow 1}\frac{a^{2}c-ac}{1+ \frac{a^{3}c^{2}}{c^2}}\\
&= \lim_{a\rightarrow 1}\frac{c\left(2a -1\right)}{3a^{2}}\\
&= \frac{c}{3}
\end{align}$$
So, it's clear that, by taking the limit in different ways, you can get an arbitrary answer. It's not valid to choose an observer moving at the speed of light and then take velocities relative to that observer.