The question uses Einstein notation: In a coordinate transformation $ x^{\mu} \rightarrow x'^{ \mu '}$ is $$ \frac{\partial ^2 x'^{ \mu '}}{\partial x^{\sigma} \partial x^{\mu}} \frac{\partial x^{\mu}}{\partial x'^{\beta '}} = \frac{\partial }{\partial x'^{\beta '}} \left[ \frac{\partial x'^{ \mu '} } {\partial x^{ \sigma} } \right]= \frac{\partial }{\partial x^{\sigma}} \left[ \frac{\partial x'^{ \mu '} } {\partial x'^{ \beta '} } \right] = 0 $$
If not why? Isn't $x' ^{\mu '}(x^{\sigma},x'^{\beta '}) \in C^2$ ?
Weirdly using this 'identity' it was somewhat easier to transform the Christoffel symbols and it gave the correct result.
It does not however give correct results under a 2-d Cartesian to polar transformation.