Since $g(f(x)) = x$ for all $x$ in the domain of $f$, we have by the Chain Rule \[g'(f(x))f'(x) = 1.\]Thus $g'(f(x)) = \dfrac{1}{f'(x)}$. We want to find $g'(0) = \dfrac{1}{f'(x)}$ for some value of $x$ such that $f(x)=0$. So we try solving $x^3-7x^2+25x-39=0$.
By the Factor Theorem, if we can find the roots of this polynomial, we can factor it. Indeed, $f(1)=0$ and $f(3)=0$, so $f(x)=(x-1)(x-3)Q(x)$ for some quadratic polynomial $Q(x)$. We set $f(x) = (x - 1)(x - 3)Q(x)$, then plug in $x = 0$ and $x = 2$ and try to solve for the coefficients of $Q(x) = ax^2 + bx + c$.
Setting $x = 0$ in the equation $f(x) = (x - 1)(x - 3)Q(x)$, we get
\[39 = (c - a)(-3).\]Setting $x = 2$ in the equation $f(x) = (x - 1)(x - 3)Q(x)$, we get
\[7 = 2a + b - 3c.\]We want to find $a, b, c$ such that these two equations hold. From the first equation, $c - a = 13$. Then $a = c - 13$. Substituting into the second equation, we get
\[7 = 2(c - 13) + b - 3c \Longrightarrow b = 5c - 33.\]So the quadratic polynomial is $Q(x) = (c - 13) x^2 + (5c - 33)x + c$. Since $f$ is cubic and has $1$ as a root, the other two roots are the roots of this polynomial. We can find those roots by using the quadratic equation formula. For example, one of the roots is \[\frac{-(5c-33) + \sqrt{(5c-33)^2-4(c-13)c}}{2(c-13)}.\]We set this equal to $3$, since we know this is one of the roots. Solving, we find $c = 3$, then $a = -10$, $b = -12$. Therefore, \[f(x) = (x - 1)(x - 3)(-10x^2 - 12x + 3).\]The derivative is given by
\begin{align*}
f'(x) &= (-10x^2 - 12x + 3) - 7(x - 3)^2 - 20x^3 - 24x^2 + 2 \\
&= -20x^3 + 35x^2 + 5x + 2,
\end{align*}and we see that $f'(1) = -10+35+5+2=32$, so \[g'(0) = \frac{1}{f'(1)} = \boxed{\frac{1}{32}}.\]
Let f(x) = x^3 - 7x^2 + 25x - 39 and let g be the inverse function of f. What is the value of g'(0)?
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