I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument on the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$

I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument in the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$

I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$ and assuming simple smoothness properties of the transformation. I found the following lovely argument on the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$

I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument on the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$