The typical translog production function $q \left(K, L\right)$ satisfies
\begin{equation}
\ln \left( q \left(K, L \right) \right) = \ln \left( A \right) + \alpha \ln \left( K \right) + \left( 1 - \alpha \right) \ln \left( L \right) + \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2
\end{equation}
with parameters $A > 0$, $\alpha \in \left( 0, 1 \right)$, $\gamma$ "close" to zero (more on that later).
To get the production function, apply the exponential:
\begin{align}
q \left(K, L \right) &= e^{\ln \left( q \left(K, L \right) \right)} \\
&= e^{\ln \left( A \right) + \alpha \ln \left( K \right) + \left( 1 - \alpha \right) \ln \left( L \right) + \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2} \\
&= e^{\ln \left( A \right)} e^{\alpha \ln \left( K \right)} e^{\left( 1 - \alpha \right) \ln \left( L \right)} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2} \\
&= A K^\alpha L^{1-\alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}
\end{align}
There are 2 elements:
- A Cobb-Douglas function: $A K^\alpha L^{1-\alpha}$
- Another term: $e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$
Let $\lambda > 1$. The objective is to analyze
\begin{equation}
q \left(\lambda K, \lambda L \right) = A \left( \lambda K \right)^\alpha \left( \lambda L \right)^{1 - \alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2}
\end{equation}
Note that
\begin{align}
\ln \left( \lambda K \right) = \ln \left( \lambda \right) + \ln \left( K \right) \\
\ln \left( \lambda L \right) = \ln \left( \lambda \right) + \ln \left( L \right)
\end{align}
and therefore
\begin{align}
\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2 &= \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda \right) + \ln \left( K \right) - \ln \left( \lambda \right) - \ln \left( L \right) \right)^2 \\
&= \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2
\end{align}
Consequently,
\begin{align}
q \left(\lambda K, \lambda L \right) &= \lambda A \left( K \right)^\alpha \left( L \right)^{1 - \alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2} \\
&= \lambda q \left(K, L \right)
\end{align}
The production function has constant returns to scale.
Note: That should not come as a surprise. The translog utility function is defined as a Taylor approximation of a CES function of the type $\left( \delta K^\gamma + \left( 1 - \delta \right) L^\gamma \right)^{\frac{1}{\gamma}}$ when $\gamma$ is close to zero. And what is a CES when $\gamma = 0$? It's a Cobb-Douglas! So that extra term in the translog ($e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$) is an error term for approximating that CES function.