Robin, you are correct about the sign mixup in the definition of the freely-falling frame. The exercise should actually say:

(Possibly Penrose forgot which set of coordinates were which: If you switch them around, the sign is correct as written - i.e. changing the sign actually gives you the
inverse coordinate transformation. However, the given expression for the wavefunction does indeed require the lower-case coordinates to be the original (non-falling) ones in order to make sense.)
You are also correct that there is a minor error in the expression for the wavefunction, however it isn't at all what your solution indicates. The error is just a missing

in the exponent of the
e. The equation in the exercise statement should actually look like this:

-----------------------------------------------------------------------------------------------
The method you used in your posted solution is a good one, however your result is
incorrect because you made an error expanding out the term

. It should be:
![{\partial_z}^2\left(e^\phi\psi\right) = e^\phi\left(\left[(\phi_z)^2+\phi_{zz}\right]\psi + 2\phi_z\psi_z + \psi_{zz}\right)](latexrender/pictures/670b2e6f1535a07453ac7b49375917e9.png)
But you omitted the

term. I re-did your calculations with this term included, and you do indeed get:

Which differs from the book only by the missing factor of

.