The coordinate transformation law (from coordinates x to coordinates y) for the Christoffel symbol is: $$\Gamma^i_{kl}(y)=\frac{\partial y^i}{\partial x^m} \frac{\partial x^n}{\partial y^k} \frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) +\frac{\partial ^2x^m}{\partial y^k \partial y^l} \frac{\partial y^i}{\partial x^m}$$ Therefore the transformation law for the contracted Christoffel symbol would be: $$\Gamma^i_{il}(y)=\frac{\partial y^i}{\partial x^m} \frac{\partial x^n}{\partial y^i} \frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) +\frac{\partial ^2x^m}{\partial y^i \partial y^l} \frac{\partial y^i}{\partial x^m}$$
$$\Gamma^i_{il}(y)= \delta^n_m\frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) + \frac{\partial }{\partial y^i }(\frac{\partial x^m}{\partial y^l}) \frac{\partial y^i}{\partial x^m}$$
$$\Gamma^i_{il}(y)= \frac{\partial x^p}{\partial y^l} \Gamma^n_{np}(x) + \frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l}) $$ My question is, does the term $\frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l})$ equal zero? My reasoning is that since partial derivatives commute, the order of differentiation can be reversed: $$\frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l})=\frac{\partial }{\partial y^l }(\frac{\partial x^m}{\partial x^m})=\frac{\partial }{\partial y^l }(\delta^m_m)=0$$ That would mean that the contracted Christoffel symbol transforms like a tensor, specifically a vector, and therefore is a tensor. However, I am unsure if it's correct to commute partial derivatives of different coordinates.
Any clarification would be greatly appreciated!