Implicit function theorem and a system of equations - am I doing this right?

Question

Show that the system of equations
$$x^2 u^2+xzv+y^2=0$$ $$yzu+xyv^2-3x=0.$$ defines implicitely the functions $u=u(x, y, z)$ and $v=v(x, y, z)$ in a neighborhood of the point $(u, v, x, y, z)=(0, 1, 3, 3, -3)$ and then compute the partial derivatives of $u(x, y, z)$ and $v(x, y, z)$ at the point $(0, 1, 3)$.
I am not sure that I did the first part correctly because I am afraid that I may be switching the variables (this is the part that really confuses me about the implicit function theorem). I considered the functions $F, G :\mathbb{R}^{5}\to \mathbb{R}$, $F(u, v, x, y, z)=x^2 u^2+xzv+y^2$ and $G(u, v, x, y, z)=yzu+xyv^2-3x$. These functions are class $C^1$ functions, $F(0, 1, 3, 3, -3)=G(0, 1, 3, 3, -3)=0$ and $\frac{D(F, G)}{D(u, v)}(0, 1, 3, 3, -3)=-81\ne 0$.
Is this what I need to do in order to reach my conclusion? I mean, can I write now that there is some open neighborhood $U$ of $(0, 1, 3)$, some open neighborhood $V$ of $(3, -3)$ and a unique pair of functions $(u, v):U \to V$, $u=u(x, y, z)$ and $v(x, y, z)$ such that $u(0, 1, 3)=v(0, 1, 3)=(3, -3)$ and $F(u(x, y, z), v(x, y, z), x, y, z)=G(u(x, y, z), v(x, y, z), x, y, z)=0$?
I am really not sure if I did the write things in order to apply the implicit function theorem. The part with the partial derivatives is easy, I just want to make sure that this is right.

Answer 1

Here's the right way to look at this. You have a function (at least $C^1$) $F\colon\Bbb R^n\to\Bbb R^k$. At some point $a\in\Bbb R^n$, suppose some $k\times k$ submatrix of $DF(a)$ is nonsingular — say the $j_1,\dots,j_k$ columns. Then you can solve locally for the variables $x_{j_1},\dots,x_{j_k}$ as $C^1$ functions of the remaining variables. You should think of this as being like the algorithm in linear algebra for solving the system $Ax=b$ by expressing the variables in pivot columns as functions of the free variables.

You can do this whenever $k$ variables give you a nonsingular "piece" of the derivative matrix. You are used to this, of course. At any point $(a_1,a_2)$ of the unit circle $x_1^2+x_2^2=1$ with neither coordinate $0$, you can locally express the circle as a graph over either the $x_1$-axis or the $x_2$-axis.