Pardon me if you find the question crazy.
How do we solve the differential equation,
$$\frac{df^{-1}(t)}{dt} \Biggm |_{f(t)} = \frac{1}{f(t)(1-f(t))}$$.
without using the following approach:
If we multiply LHS and RHS by $df$, by inverse function theorem we have for LHS, $$ \frac{df^{-1}(t)}{dt} \Biggm |_{f(t)} df = dt $$
And thus,
$$ \frac{df}{dt} = f(t)(1-f(t))$$ and we can solve this.
Notice that the notation $$\frac{df^{-1}(t)}{dt}\lvert_{f(t)}$$ means $$\frac{df^{-1}(f(t))}{df(t)}=\frac{dt}{df(t)}$$
So we obtain $$\frac{dt}{df}=\frac{1}{f\cdot(1-f)}$$ which is separable: $$dt=\frac{df}{f\cdot(1-f)}\implies t+C=\ln|f\cdot(1-f)|$$