My attempt:
I tried to use the chain rule to prove that $ h(t) = xe^{f(t)} $ implies that $h(t) =x + \int_{[0,t]}h(s) df(s)$ but it was fruitless. To prove the other way I tried to prove that $h(t)exp\{-f(t)\} $ is constant w.r.t $t$, for that purpose I tried to derive $h(t)exp\{-f(t)\} $ w.r.t. f (in the radon nikodyn sense). But I can't formalize that \begin{align*} \frac{d}{df}h(t)exp\{-f(t)\} & = h(t)*\frac{d}{df} exp\{-f(t)\} + exp\{-f(t)\}*\frac{dh(t)}{df }\\ &= -h(t) exp\{-f(t)\} + exp\{-f(t)\}h(t)\\ &= 0. \end{align*} I will appreciate any help.
The question has been edited after I posted this.
If $f$ is continuously differentiable then $h(t)=x+\int_0^{t} h(s)f'(s)ds$ and so $h'(t)=h(t)f(t)$. This gives $h(t)=c e^{f(t)}$. But $h(0)=x$ so we get $h(t)=xe^{f(t)}$.