Partial Derivatives: Chain rule

[Baron]

The question is in the attachments and my attempt is also in the attachments.

I'm kind of stuck. I'm not sure whether I took the second partial derivative correctly or how to simplify
(∂^2 G)/(∂x ∂b)

I have no idea how I'm going to use the fact ∂F/∂z = (∂^2 F)/(∂x^2) + (∂^2 F)/(∂y^2)

 

[HallsofIvy]

You are given that F(x, y, z)= G(b, s, t) so \(\displaystyle \dfrac{\partial F}{\partial z}= \dfrac{\partial G}{\partial t}\), \(\displaystyle \dfrac{\partial^2F}{\partial x^2}= \dfrac{\partial^2 G}{\partial b^2}\), and \(\displaystyle \dfrac{\partial^2 F}{\partial y^2}= \dfrac{\partial^2 G}{\partial s^2}\). That is, \(\displaystyle \dfrac{\partial F}{\partial z}= \dfrac{\partial^2 F}{\partial x^2}+ \dfrac{\partial^2 F}{\partial y^2}\) becomes \(\displaystyle \dfrac{\partial G}{\partial t}= \dfrac{\partial^2 G}{\partial b^2}+ \dfrac{\partial^2 G}{\partial s^2}\)

 

[Baron]

You are given that F(x, y, z)= G(b, s, t) so \(\displaystyle \dfrac{\partial F}{\partial z}= \dfrac{\partial G}{\partial t}\), \(\displaystyle \dfrac{\partial^2F}{\partial x^2}= \dfrac{\partial^2 G}{\partial b^2}\), and \(\displaystyle \dfrac{\partial^2 F}{\partial y^2}= \dfrac{\partial^2 G}{\partial s^2}\). That is,
\(\displaystyle \dfrac{\partial F}{\partial z}= \dfrac{\partial^2 F}{\partial x^2}+ \dfrac{\partial^2 F}{\partial y^2}\) becomes \(\displaystyle \dfrac{\partial G}{\partial t}= \dfrac{\partial^2 G}{\partial b^2}+ \dfrac{\partial^2 G}{\partial s^2}\)

If G(b,s,t) stasifies the equation \(\displaystyle \dfrac{\partial G}{\partial t}= \dfrac{\partial^2 G}{\partial b^2}+ \dfrac{\partial^2 G}{\partial s^2}\), for all values, does that mean the value of A can be any number?