$dS_t = \mu_tS_tdt+\sigma_t S_tdB_t$, if define $X(t)=\int_0^t \mu(s)-1/2\sigma^2(s)ds + \int_0^t \sigma(s)dB(s）$,then the solution of S is $S_t = S_0 \exp(X_t)$.
now define $A_t = \int_0^tS_udu$, $Y_t = A_t/S_t$.
prove:$dY_t = (1-\mu_tY_t)dt + \sigma_tY_tdB_t$
I saw this proposition in a paper without proof, but I cannot prove it myself, especially use ito to $A_t$, I haven't calculated the problem of this nested form before.
On
Another, more direct, way to approach the problem is to Ito the multidimensional Ito formula.
Consider the function $f(x,y):=x/y$ and apply Ito in two dimensions, i.e.
$$ df = \bigg( \partial_t f + \mu_x \partial_xf+\mu_y\partial_y f+\frac{\sigma_{x}^{2}}{2}\partial_{xx}f +\frac{\sigma_{y}^{2}}{2}\partial_{yy}f + \sigma_x \sigma_y \rho \, \partial_{xy} f\bigg)dt + \sigma_x \partial_xf dB_{t}^{1} + \sigma_y \partial_yf dB_{t}^{2} $$
where $B^i$ are two correlated BM with correlation $\rho$.
In our case then
$$ dY_t=(1+(\sigma_{t}^{2} - \mu_t)Y_t)dt - \sigma_t Y_t dB_t $$
You have $dA_t=S_t\,dt$.
Division can look ugly, look at the multiplication $A_t=S_tY_t$, with the trial $dY_t=adt+bdB$. Then \begin{align} S_t\,dt=dA_t&=Y_tdS_t+S_tdY_t+d\langle S,Y\rangle_t \\ &=S_tY_t(μ_t\,dt+σ_t\,dB_t)+S_t(a_t\,dt+b_t\,dB_t)+S_tσ_t\,b_t\,dt \\ dt &= (μ_tY_t+a_t+σ_t\,b_t)\,dt + (σ_tY_t+b_t)\,dB_t \end{align}
Comparing coefficients gives $b_t=-σ_tY_t$ and $a_t=1-μ_tY_t+σ_t^2Y_t$.
Let's try alternatively for, $Y_t=S_t/A_t$.
Then repeating the same as above gives for $S=AY$ \begin{align} S_t(μ_t\,dt+σ_t\,dB_t)=dS_t&=A_tdY_t+Y_tdA_t+d⟨A,Y⟩_t \\ &=A_t(a_tdt+b_tdB_t)+Y_tS_t\,dt+0 \\ Y_t(μ_t\,dt+σ_t\,dB_t)&=(a_t+Y_t^2)\,dt+b_t\,dB_t \end{align} so that $a_t=Y_tμ_t-Y_t^2=Y_t(μ_t-Y_t)$ and $b_t=σ_tY_t$.
This gives the SDE $$ dY_t=Y_t(μ_t-Y_t)\,dt+σ_tY_t\,dB_t, $$ still not the claimed equation.