For $G:\mathbb{R^3} \to \mathbb{R}$, and $F:\mathbb{R^2} \to \mathbb{R}$ show $G(x,y,z)=F(x+2y+3z-1,x^3+y^2-z^2)=0$ can have $z=g(x,y)$

Question

For the problem below from Munkres' Analysis on Manifolds, I am having trouble showing that we can solve $G(x,y,z)=0$ for $z=g(x,y)$ in a neighborhood of $(-2,3)$

I know that if $G$ were given explicitly then by the implici function theorom, I'd just have to check that $\frac{\partial G}{\partial z} \neq 0$. But since $G$ is defined in terms of $f$, I'm not sure how to check this. Would it be equivalent to checking $\frac{\partial F}{\partial z} \neq 0$? Even then I'm not sure how to proceed.

Thanks

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Answer 1

Calculation of $\frac{\partial G}{\partial z}$ is just an application of the chain rule.

If you are not used to it, to avoid confusion you should first name the parameters of $F$, like $F= F(u, v)$

Then $u, v$ must be written as functions of $x, y, z$ (like $u= x + 2y + 3z -1$) and you have to calculate $$ \frac{\partial G}{\partial z} = \frac{\partial F}{\partial u} \frac{\partial u}{\partial z} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial z}$$

The actual calculation I leave to you.