This is very straightforward, just substitute the transformation rules and collect the terms.
Here are some details.
The inverse metric transforms, as we know, by the rule: $ g^{\mu \lambda} = \frac{\partial{\bar{x}}^\mu}{\partial{x}^\alpha} \frac{\partial{\bar{x}}^\lambda}{\partial{x}^\delta} g^{\alpha \delta} $
The partial derivatives need some calculations that can be presented as $ \begin{align*} g_{\lambda \kappa , \nu} & = \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\delta \gamma} \Big) \\ &= \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} g_{\delta \gamma , \beta} + g_{\delta \gamma} \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big) \end{align*} $
Similarly, $ g_{\nu \lambda , \kappa} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\beta \delta , \gamma} + g_{\beta \delta} \frac{\partial}{\partial{\bar{x}^\kappa}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \Big) $ and $ g_{\nu \kappa , \lambda} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} g_{\beta \gamma , \delta} + g_{\beta \gamma} \frac{\partial}{\partial{\bar{x}^\lambda}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big) $
Substituting these identities into your "definition" $ \Gamma^\mu _{\nu\kappa} = \frac{1}{2}g^{\mu\lambda}\left(g_{\lambda\kappa,\nu}+g_{\nu\lambda,\kappa}-g_{\nu\kappa,\lambda} \right) $ and taking into account that $ \Gamma^\alpha _{\beta \gamma} = \frac{1}{2}g^{\alpha \delta}\left(g_{\delta \gamma , \beta}+g_{\beta \delta , \gamma} - g_{\beta \gamma , \delta} \right) $ it is not difficult now to show the required transformation rule for the Christoffel symbols.