For two functions to be inverses, this must be true:
$$ f(g(x))=x $$
We can differentiate this equation to obtain the relationship between the derivatives of the function.
$$ f'(g(x))\cdot g'(x) = 1 $$
For this question, we are trying to find \(g'(1)\).
$$ g'(1)= \frac{1}{f'(g(1))} $$
Since the question mentions that \(f(0)=1\), \(g(1)=0\) must be true since \(f\) and \(g\) are inverses.
$$ g'(1)=\frac{1}{f'(0)} $$
Find \(f'(0)\):
$$ f(x)=(2x+1)^3 $$
$$ f'(x)=6(2x+1)^2 $$
$$ f'(0)=6 $$
Putting it all together:
$$ g'(1)=\boxed{\frac{1}{6}}$$
We can swap \(x\) and \(y\) values to obtain the equation of the inverse function.
$$ y=(2x+1)^3 $$
$$ x=(2y+1)^3 $$
Solve for \(y\):
$$ \sqrt[3]{x}=2y+1 $$
$$ y = \frac{\sqrt[3]{x}}{2}-\frac{1}{2} $$
The inverse function \(g\) can be differentiated:
$$ g(x)=\frac{1}{2}x^{\frac{1}{3}}-\frac{1}{2} $$
$$ g'(x)=\frac{1}{6}x^{-\frac{2}{3}} $$
$$ g'(1) = \frac{1}{6}(1)^{-\frac{2}{3}} $$
$$ g'(1)= \boxed{\frac{1}{6}} $$