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Honestly, I don't understand the question well. As a try, I defined a function $G:\mathbb{R}^{2+1}\rightarrow \mathbb{R}^2$ where $G(x,y,z)=(f_1(x,y,z),f_2(x,y,z))$ so that $f_1(x,y,z)=x-y$ and $f_2(x,y,z)=y-z$. Then, to satisfy the condition of Implicit function theorem, we need $G(a,b,c)=(0,0)$ for some $(a,b,c)\in \mathbb{R}^2$. That happens when $a=b=c$. So, I think I am done for the first part but not sure. Am I correct? Secondly, how may I derive those explicit formulas? Any help is appreciated.

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Hint.

If you set $$G(x,y,z)=F(x-y,y-z)$$ you need to look at points $(a,b,c)$ where $$G(a,b,c)=F(a-b,b-c)=0.$$ You also need to have at such points $\frac{\partial G}{\partial z}\neq 0$ and using the chain rule you have $$\frac{\partial G}{\partial z}(x,y,z) = -\frac{\partial F}{\partial x_2}(x,y,z).$$ Finally you get the two conditions $$\begin{cases} F(a-b,b-c)=0\\ \frac{\partial F}{\partial x_2}(a-b,b-c) \neq 0 \end{cases}$$ To compute the partial derivatives of $z(x,y)$ you have to use the chain rule.

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