How does the Wavefunction change after constant to the potential energy? My question is related to this older question.
I am currently working my way through what appears to be a newer edition of Griffiths: "Introduction to Quantum Mechanics" book. I am stumped by how to tackle problem 1.8, which states:

Suppose you add a constant $V_0$ to the potential energy (by “constant” I mean
independent of $x$ as well as $t$). In classical mechanics this doesn’t change anything, but
what about quantum mechanics? Show that the wave function picks up a time-dependent
phase factor: $e^{−iV_0t/\hbar}$. What effect does this have on the expectation value of a
dynamical variable?

In the linked physic.se question the questioner outlines a method for solving this problem. I am going to echo that method, with slightly different notation. He begins with the Schödingers equation,
$$
\bbox[.5em, border:.1em solid gray]{
i\hbar\frac{\partial \Psi}{\partial t} = \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} +  V \Psi}
$$
He adds the constant $V_0$ to get,
$$
i\hbar\frac{\partial}{\partial t} \Psi(x,t) = 
\left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) + V_0\right ] \Psi(x,t).
$$
At this point he does something strange to me, and asking over at math.se chat I learned that this trick is probably separation of variables.
I think what he does is, assume $\Psi(x,t)$ can be approximated by, $f(x)g(t)$,
then we can rewrite Schrödigner as
$$
i\hbar\frac{\partial}{\partial t} f(x)g(t) = \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) + V_0\right ] f(x)g(t).
$$
He then claims to divide by $f(x)g(t)$ to get
$$
\frac{\partial g(t)}{\partial t} = \frac{-i}{\hbar}(E+V_0)t.
$$
However when I try to make the same move, I am left with, $$
\frac{i\hbar\frac{\partial}{\partial t} f(x)g(t)}{f(x)g(t)} = \frac{1}{f(x)g(t)}\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}f(x)g(t) + V(x,t) + V_0
\\ \implies \\
\frac{i\hbar\frac{\partial}{\partial t}g(t)}{g(t)} = \frac{1}{f(x)}\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}f(x) + V(x,t) + V_0
$$
What am I missing?

For reference, regarding what I understand, my maths is pretty low-level, I took a course in undergrad calculus over 10 years ago and only just passed that at the time, and my physics knowledge is not much further than that of an enthusiastic lay person.
 A: You are missing the fact, that it is assumed that $f(x)$ is an energy eigenfunction with energy eigenvalue $E$ that satisfies the time independent Schrödinger equation,
$$
\left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ] f(x) = Ef(x),
$$ where I also assume that your potential is NOT a function of time, and merely a function of position $V=V(x)\neq V(x,t)$. With these assumptions we can proceed as follows,
$$\begin{aligned}
i\hbar\frac{\partial}{\partial t} f(x)g(t) &= \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) + V_0\right ] f(x)g(t)\\
f(x)i\hbar\frac{\partial}{\partial t} g(t)&=g(t)\left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) + V_0\right ] f(x)\\
&=g(t)\left( \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right]f(x) + V_0f(x)\right) \\
&=g(t)\left(  Ef(x) + V_0f(x)\right) \\
f(x)i\hbar\frac{\partial}{\partial t} g(t)&=g(t)( E +V_0)f(x) \quad |\text{Divide by }f(x) \\
i\hbar\frac{\partial}{\partial t} g(t)&=g(t)( E +V_0) \\
\frac{1}{g(t)}\frac{\partial}{\partial t} g(t)&=-i( E +V_0)/\hbar  \quad | \text{Integrate both sides with respect to t and solve for g(t)}\\
g(t) &=e^{-i(E+V_0)t/\hbar } 
\end{aligned}$$
As you can see, the time dependency is simply a phase factor whose angular frequency is determined by its energy $\omega = E/\hbar$. The addition of a constant potential shifts this angular frequency $\omega \rightarrow \omega'  = (E+V_0)/\hbar$  just as it shifts the energy eigenvalue $E\rightarrow E+V_0$. The function $f(x)$ remains the same function meanwhile, only its associated eigenvalue does change by the constant value of the added potential.
