= Solution
Since $H$ is continuous, strictly increasing and unbounded, $J_t=\inf\{s>0:H_s>t\}$ is its continuous inverse: $H_{J_t}=t$, $J_{H_s}=s$ and $J_0=0$. These inverse times are <stopping times>. Put $X_t=Y_{J_t}$, which is continuous and strictly positive and is adapted to the inverse-clock filtration $\mathcal G_t=\mathcal F_{J_t}$.
Evaluate the identity from part (a) at $J_t$. The Lebesgue-Stieltjes change of variable $r=H_s$ gives
$$
\int_0^{J_t}\frac1{Y_s}\,dH_s=\int_0^{H_{J_t}}\frac1{Y_{J_r}}dr=\int_0^t\frac1{X_r}dr.
$$
Therefore
$$
\boxed{X_t=1+\beta_t+\int_0^t\frac1{X_s}ds,\qquad dX_t=d\beta_t+\frac1{X_t}dt.}
$$
The process $\beta$ is <Brownian motion> in this same filtration by the construction in part (a). The drift integral is finite on each compact interval because the positive continuous process $X$ has a positive minimum there. This is the dimension-three case of the <Exponential Brownian-to-Bessel time change>.
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