Let .
Then
That's the third derivate.
$\displaystyle f(x)=\frac{1-4x}{1+4x}=\frac{2}{1+4x}-\frac{1+4x}{1+4x}=\frac{2}{1+4x}-1$$\displaystyle \Rightarrow f'''(x)=-\frac{2\cdot 4^3 \cdot 2 \cdot 3}{(1+4x)^4}=-\frac{768}{(1+4x)^4}$ and replace with $\displaystyle x=5$.