Let $\mathcal C$ be the locus of all points $(x,y,z)$ in $\mathbb R^3$ that satisfy both $$x^3(y^3+z^3)=0\text{ and } (x-y)^3=z^2+7.$$ Prove that near the point $p:=(1,-1,1)$, we can solve for $y$ and $z$ as smooth functions of $x$. That is, show that there exists an $r>0$ and $\mathcal C^1$ functions $f,g:(1-r,1+r) \rightarrow \mathbb R$ such that $p=(1,f(1),g(1))$ and for all $x \in (1-r,1+r)$, $(x, f(x),g(x))$ lies on $\mathcal C$.
I know that I will need to employ the Implicit Function Theorem, but I'm not sure how to construct a function given the two equations. Also will I need to apply it separately to get $f$ and $g$ or can I get a function $h: \mathbb R^2 \rightarrow \mathbb R$?
This is a trick you can only use in $\mathbb{R}$. Suppose you want to solve two real valued functions simultaneously for roots, I.E. We have some $f,g$ and we want $f(x)=0$ and $g(x)=0$. Then we might as well solve $$ f(x)^2 + g(x)^2=0$$ Because in the reals, a sum of squares is $0$ if and only if each term is $0$.
Use this to reduce your problem to just a single equation and then use implicit function theorem to finish it off. Here, your two functions will have 3 parameters, but that's just notation.