I think I know what is going on but please point out my mistakes and gaps in understanding!
1 the author decomposes the partial using CHAIN RULE
$$\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}$$
2 the author uses the PRODUCT RULE somehow to get this?
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial \psi}{\partial r}\frac{\partial^2 r}{\partial x^2}$$
how does he use the product rule?: I am guessing he assigns:
$$u = \frac{\partial \psi}{\partial r}$$ $$v = \frac{\partial r}{\partial x}$$ $$\frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial x}\right) = \frac{\partial}{\partial x}\left( \frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right) = u \frac{\partial}{\partial x}\left(v\right)+v \frac{\partial}{\partial x}\left(u\right)$$
$$ \frac{\partial \psi}{\partial r}\times \frac{\partial}{\partial x}\left(\frac{\partial r}{\partial x}\right) + \frac{\partial r}{\partial x}\times \frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial r}\right) $$
I do not understand how the author turned $$ \frac{\partial \psi}{\partial r}\times \frac{\partial}{\partial x}\left(\frac{\partial r}{\partial x}\right)$$
into
$$ \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 $$
my understanding now from reading the answers is that
$$\frac{\partial}{\partial x} \frac{\partial \psi}{\partial r}= \frac{\partial^2}{\partial x \partial r } $$
$$\frac{\partial^2 \psi}{\partial x \partial r } = \frac{\partial}{\partial r}\frac{\partial \psi}{\partial x}$$
but $$\frac{\partial \psi}{\partial x} = \left(\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right)$$
so $$\frac{\partial}{\partial r} \left(\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right)$$
$$\frac{\partial^2 \psi}{\partial r^2}\frac{\partial r}{\partial x}$$
multiplying this by the $$\frac{\partial r}{\partial x}$$ that I left out
results in $$ \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 $$
could I start an infinite loop by factoring out another $$\frac{\partial \psi}{\partial x}$$ and replace it with $$\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}$$
$$ \frac{\partial^2 \psi}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial \psi}{\partial r} \frac{\partial r}{\partial x} \right) = \frac{\partial^2 \psi}{\partial x \partial r}\frac{\partial r}{\partial x} + \frac{\partial \psi}{\partial r} \frac{\partial^2 r}{\partial x^2} = \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial \psi}{\partial r} \frac{\partial^2 r}{\partial x^2} $$
The first equality is the chain rule, the second is the product rule, then the last one is the chain rule again. Do you see how the last equality follows from the chain rule?
Edit: $$\frac{\partial^2 \psi}{\partial x \partial r} = \frac{\partial}{\partial r}\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial r} \left(\frac{\partial \psi }{\partial r}\frac{\partial r}{\partial x}\right) = \frac{\partial^2 \psi}{\partial r^2} \frac{\partial x}{\partial r} +\frac{\partial \psi}{\partial r}\frac{\partial}{\partial x} \frac{\partial r}{\partial r} $$ The last term is equal to zero.