I am studying G.R an I have a mistake by using chain rules (yes probably very basic mistake) that I don't get.
Take the following quantity (there is Einstein convention here, the $X$ and $V$ are vectors and $\{x\}$ and $\{y\}$ are two different system of coordinates.):
$$ X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})$$
I need to "simplify" it.
I can do it in two different ways.
First, I can use the summation over $\mu$ :
$$ \frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}=\frac{\partial x^{\alpha}}{\partial x^{\beta}}=\delta^{\alpha}_{\beta} $$
I also could use the summation over $\alpha$ (other "method" to simplify it):
$$\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha}x^{\sigma}}=\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial}{\partial x^{\alpha}}\frac{\partial y^{\rho}}{\partial x^{\sigma}}=\frac{\partial y^{\rho}}{\partial y^{\mu} \partial x^{\sigma}}=\frac{\partial}{\partial x^{\sigma}}\frac{\partial y^{\rho}}{\partial y^{\mu}}=\frac{\partial}{\partial x^{\sigma}}\delta^{\rho}_{\mu}=0$$
Thus, if I use the first thing I will end up with :
$$ X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})=X^{\beta} V^{\sigma} (\frac{\partial^2 y^{\rho}}{\partial x^{\beta} \partial x^{\sigma}}) \neq 0$$
And if I use the second method I end up with :
$$ X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})=0$$
It is probably a very stupid mistake or a more deep incomprehension thing but I really don't see it.
Could you help me figure this out ?