Use the implicit function theorem to discuss the solvability of the system $$3x + 2y + z^2 + u + v^2 =0 \\ 4x+ 3y+ z + u^2 + v + w + 2 = 0\\ x + z + u^2 + w + 2 = 0\\$$ for $u , v, w$ in terms of $x, y, z$ near $x = y = z = 0$ and $u = v= 0$ and $w = 2$.
In approaching this question, I'm quite stuck as usually questions on implicit function theorem deal with three variables, but this deals with 6. Should I be looking at a matrix that has 6 columns? If so, how would I continue to solve this system using the implicit function theorem?
First there appears to be a typo in your post. The $w$ should be $-2$. To this end, define the $3$ $C^{1}$ functions: $F_1 = 3x+2y+z^2+u+v^2, F_2 = 4x+3y+z+u^2+v+w+2, F_3 = x+z + u^2+w+2 $. We calculate the determinant at point $(0,0,0,0,0,-2)$ and show it not equal to $0$.
$\text{det}\begin{bmatrix} \dfrac{\partial F_1}{\partial u} & \dfrac{\partial F_1}{\partial v} & \dfrac{\partial F_1}{\partial w} \\ \dfrac{\partial F_2}{\partial u} & \dfrac{\partial F_2}{\partial v} & \dfrac{\partial F_2}{\partial w} \\ \dfrac{\partial F_3}{\partial u} & \dfrac{\partial F_3}{\partial v} & \dfrac{\partial F_3}{\partial w} \end{bmatrix}|_{(0,0,0,0,0,-2)} = \text{det}\begin{bmatrix} 1 & 2v & 0 \\ 2u & 1 & 1 \\ 2u & 0 & 1 \end{bmatrix}|_{(0,0,0,0,0,-2)}= 1 \neq 0 $. Thus, you can write $u = f(x,y,z), v = g(x,y,z), w = h(x,y,z)$ near the point $(x,y,z,u,v,w) = (0,0,0,0,0,-2)$.