This is my guess:
We start with the following equation:
$$
\pi_t = \frac{(1-\mu)(1-\lambda)}{\mu}\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right]^{(\sigma-\rho)/\rho}\left(\frac{h_{ut}}{h_{st}}\right)^{(1-\sigma)}\left(\frac{\psi_{st}}{\psi_{ut}}\right)^\sigma
$$
Taking logs gives:
$$
\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) + \frac{\sigma - \rho}{\rho}\ln\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right] + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}
$$
Take the approximation $\ln(1 + x ) \approx x$ to simplify the second term to:
$$
\frac{\sigma - \rho}{\rho} \left[-\lambda + \lambda \left(\frac{k_{et}}{s_t}\right)^\rho\right]
= -\lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho
$$
This gives:
$$
\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) - \lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho\\ + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}
$$
Now differatiating with respect to time gives on the left hand side $g_{\pi_t}$. The first two terms on the right hand side disappear. For the third term we get:
$$
\begin{align*}
&\frac{\sigma - \rho}{\rho} \lambda \rho \left(\frac{k_{et}}{s_t}\right)^{\rho - 1} \frac{\dot k_t s_t - k_t \dot s_t}{(s_t)^2},\\
&= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho \frac{s_t}{k_t} \frac{\dot k_t - k_t g_{s_t}}{s_t},\\
&= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho (g_{k_t} - g_{s_t})
\end{align*}
$$
I think $g_{s_t} = g_{h_{st}} + g_{\psi_{st}}$. The other terms give:
$$
(1-\sigma) (g_{h_{ut}} - g_{h_{st}}) + \sigma (g_{\psi_{st}} - g_{\psi_{ut}})
$$