Let's first make some substitution. Define $g = f^\varepsilon s$. This gives:
$$
\begin{align*}
L &= \frac{\alpha g[1 - Ak(1 - \alpha + \alpha g)^{1/\varepsilon} g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&= \frac{\alpha g[1 - AK(1 - \alpha(1 - g))^{1/\varepsilon}g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&= \frac{\alpha \left[1 - Ak\left(\frac{1- \alpha(1 - g)}{g}\right)^{1/\varepsilon}\right]}{\frac{1 - \alpha(1+g)}{\alpha g}}
\end{align*}
$$
The two terms on $\alpha$ and $g$ in the denominator and numerator are almost the same except from a sign reverseal. So maybe this is a typo. I will continue by assuming that it is.
Assume that the signs on $g$ were flipped, then define
$$
h = \frac{1 + \alpha(1- g)}{g}
$$
This allows the condensed form:
$$
L = \frac{\alpha[1 - Ak h^{1/\varepsilon}]}{h} \tag{1}
$$
Then:
$$
\ln L = \ln \alpha + \ln(1 - A k h^{1/\varepsilon}) - \ln h
$$
And:
$$
\frac{\partial \ln L}{\partial t} = \frac{\partial \ln(1 - A k h^{1/\varepsilon})}{\partial t} - \frac{\partial \ln h}{\partial t}
$$
The first derivative on the right hand side equals:
$$
\begin{align*}
\frac{\partial (1 - A k h^{1/\varepsilon})}{\partial t} &=
-\frac{A k h^{1/\varepsilon} \dfrac{\dot k}{k} + \frac{1}{\varepsilon} A k h^{1/\varepsilon}\dfrac{\dot h}{h}}{1 - A k h^{1/\varepsilon}},\\
&=- \frac{A h^{1/\varepsilon}k}{1 - A k h^{1/\varepsilon}}\frac{\dot k}{k}- \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon} }{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h}
\end{align*}
$$
So:
$$
\frac{\dot L}{L} = - \frac{A k h^{1/\varepsilon}}{1 - Ak h^{1/\varepsilon}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon}}{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h} - \frac{\dot h}{h}
$$
Now we know from $(1)$ that:
$$
1 - A k h^{1/\varepsilon} = \frac{L h}{\alpha},
$$
So:
$$
\frac{\dot L}{L} = - \dfrac{1-\dfrac{L h}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{1-\dfrac{Lh}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot h}{h}- \frac{\dot h}{h},\\
= - \left(\frac{\alpha}{Lh} - 1\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1\right)\frac{\dot h}{h} - \frac{\dot h}{h},\\
= \left(1- \frac{\alpha}{Lh}\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{\dot h}{h}
$$
Next, $h = \dfrac{1 + \alpha(1 -g)}{g}$ so taking logs gives:
$$
\ln h = \ln (1 + \alpha(1 - g)) - \ln g
$$
taking the derivative with respect to time gives:
$$
\frac{\dot h}{h} = \frac{-\alpha g}{1 + \alpha(1 - g)}\frac{\dot g}{g} - \frac{\dot g}{g},\\
= - \frac{1 + \alpha}{1 + \alpha( 1 - g)}\frac{\dot g}{g},\\
= -\frac{(1+\alpha)}{hg} \frac{\dot g}{g},\\
= -\frac{hg + \alpha g}{hg} \frac{\dot g}{g},\\
= \frac{h + \alpha}{h} \frac{\dot g}{g}
$$
Finally, $g = f^\varepsilon s$ so:
$$
\frac{\dot g}{g} = \varepsilon \frac{\dot f}{f} + \frac{\dot s}{s}
$$
Together this gives:
$$
\begin{align*}
\frac{\dot L}{L} &= \left(1 - \frac{\alpha}{Lh}\right) \frac{\dot k}{k} ,\\
&+ \left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot f}{f},\\
&+\frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot s}{s}
\end{align*}
$$
