We are asked to show that there's a neighborhood $U$ of the point $(3,0)$ such that for any $(a,b) \in U$ there's a solution to the system $\begin{cases}x^3+y^3+z^3=a \\ xy-z^2 = b\end{cases}$
What I did:
Notice that if $(a,b) = (3,0)$ then $(x,y,z) = (1,1,1)$ is a possible solution. Define the function $F(x,y,z,a,b) = \begin{pmatrix}x^3 + y^3+z^3 - a \\ xy-z^2 - b \end{pmatrix}$.
Notice that $F \in C^1$ and that $F(1,1,1,3,0) = \begin{pmatrix}0\\0\end{pmatrix}$.
$D_F(x,y,z,a,b) = \begin{pmatrix}3x^2 & 3y^2 & 3z^2 & -1 & 0\\y & x & -2z & 0 & -1\end{pmatrix} =_{(1,1,1,3,0)}\begin{pmatrix}3 & 3 & 3 & -1 & 0 \\ 1& 1 & -2 & 0 & -1\end{pmatrix}$
The sub matrix $\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ is invertible, so from the implicit function theorem there is an open set $V$, $(1,1,1) \in V$, and an open set $U$, $(3,0) \in U$, and a $C^1$ function $g:V \to U$ such that $g(1,1,1) = (3,0)$ and $\forall (x,y,z) \in V: F(x,y,z,g(x,y,z)) = (0,0)$
But that's not really what we wanted to show. If anything, I've shown that for any $(x,y,z)$ in a neighborhood of $(1,1,1)$ there's a solution. Not that for any $(a,b)$ in a neighborhood of $(3,0)$ there's a solution.
Am I confused?
You're confused, what you did is correct. It might help to note that you're using $(a,b)$ with two different meanings: one is the problem statement, the other in the solution.
Rewrite the beginning of your solution as
Continue with your solution and conclude that there is an open set $U$ and a $C^1$ function $g:V \to U$ such that:
Now recall your goal: to prove that there exists a neighborhood of $(3,0)$, say, $W$ such that for any $(a,b)$ in $W$ $$\exists x,y,z\in \mathbb R\left(F(x,y,z,a,b)=(0,0)\right)_.$$
Set $W=g[V]$ (depending on how you state and prove the implicit function theorem, you can actually let $W$ be $U$). Now let $(a,b)$ be any element of $W$. Since $(a,b)$ belongs to $g[V]$, there exist $x,y,z$ in $V$ such that $g(x,y,z)=(a,b)$. Thus $F(x,y,z,a,b)=(0,0)$.