Solving a 5 dimensional function in a neighbourhood Consider a function $f:\mathbb{R}^5 \to \mathbb{R}^2$ defined by
$$f(u,v,w,x,y)=(uy+vx+w+x^2,uvw+x+y+1)$$
such that $f(2,1,0,-1,0)=(0,0)$
(i) Show that we can solve $f(u,v,w,x,y) = (0,0)$ for $(x,y)$ in terms of (u,v,w) in a neighbourhood of $(2,1,0)$.
(ii)If $(x,y) = \phi(u,v,w)$ is the solution for (i) then show that derivative of $\phi$ at $(2,1,0)$ is
$$D\phi(2,1,0)=\frac13
\begin{bmatrix}
        0 & -1 & -3 \\
        0 & 1 & -3 \\
        \end{bmatrix}$$
Here's how I tried:
Let $F=uy+vx+w+x^2=0$
& $G=uvw+x+y+1=0$
$$\frac{\partial(F,G)}{\partial(x,y)}_{(2,1,0,-1,0)} =
\begin{bmatrix}
        v+2x & u \\
        1 & 1 \\
        \end{bmatrix}$$
$$\qquad \qquad \qquad=
\begin{bmatrix}
        -1 & 2 \\
        1 & 1 \\
        \end{bmatrix}$$
which is non singular, so solution exists.
Part (ii):
we can write 
$$x=X(u,v,w)$$
$$y=Y(u,v,w)$$
defined in the neighbourhood of $(2,1,0)$
such that:
$$X(2,1,0)=-1$$
$$Y(2,1,0)=0$$
what to do next? How to find $x=X(u,v,w)$ & $y=Y(u,v,w)$?
 A: The solution need not be nearly as complicated as what izœc does.
To show that $\phi$ exists in a neighborhood of $(2,1,0)$, we do exactly as you did:
$$Df(u,v,w,x,y) = \begin{pmatrix} y & x & 1 & v + 2x & u \\ vw & uw & uv & 1 & 1 \end{pmatrix},$$
$$Df(2,1,0,-1,0) = \begin{pmatrix} 0 & -1 & 1 & -1 & 2 \\ 0 & 0 & 2 & 1 & 1 \end{pmatrix}.$$
Since the matrix
$$\begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix}$$
corresponding to $(x,y)$ is invertible, the implicit function theorem can be applied.
Now if we write $Df = (A | B)$, where
$$A = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix},$$
a little work with the chain rule shows us that
$$D\phi(2,1,0) = -A^{-1} B.$$
This works for all implicit function theorem problems; see Rudin's Principles of Mathematical Analysis for the details. In our case,
$$A^{-1} = -\frac{1}{3} \begin{pmatrix} 1 & -2 \\ -1 & -1 \end{pmatrix},$$
and hence
\begin{align}
D\phi(2,1,0) & = \frac{1}{3} \begin{pmatrix} 1 & -2 \\\ -1 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & 2 \end{pmatrix} \\
 & = \frac{1}{3} \begin{pmatrix} 0 & -1 & -3 \\ 0 & 1 & -3 \end{pmatrix}. 
\end{align}
