showing that system of equations can be solved using the implicit function theorem I have to show that $xy^5 + yu^5 + zv^5 = 1, x^5y + y^5u + z^5v = 1$ can be solved in a neighbourhood of $(0,1,1,1,0)$. I can show that the two equations are solved if and only if $u=f(x,y,z,v)$ and $v=g(x,y,z,u)$ in a neighbourhood of $(0,1,1,1,0)$. My problem now is that I can't figure out how to make f and g only depend on $(x,y,z)$.
 A: $\bullet$  Let $M =\{xy^5 + yu^5 + zv^5 = 1\} \cap \{ x^5y + y^5u + z^5v = 1\}$
$\bullet$ Let $F(x,y,z,u,v) = (xy^5 + yu^5 + zv^5 -1,x^5y + y^5u + z^5v-1)= (r,s)$
$\bullet$ Then $M = F^{-1}(0,0)$ and $F$ is smooth. 
$\bullet$ Letting $p = (x,y,z,u,v)$ we have,
$$DF(p) = \begin{pmatrix} y^5 & 5xy^4 + u^5 & v^5 & 5yu^4 & 5zv^4 \\ 5x^4y & x^5+5y^4u & 5z^4v & y^5 & z^5  \end{pmatrix}$$
$\bullet$ Show that this matrix has rank $2$, which will imply by the level-set theorem, that $M$ is a smooth manifold of dimension $3$. 
$\bullet$ Hence we can allow $(u,v)$ to be our dependent variables. Since $M$ is a manifold, and $(x,y,z,u,v)$ give the coordinates on $M$, we must have that each coordinate is smooth i.e $u = g(x,y,z)$ and $v = h(x,y,z)$ where $h,g$ are smooth. To show that this is in fact true, we use the inverse function theorem on the map,
$$ T(\textbf{x}) = \begin{pmatrix}  \frac{\partial F}{\partial u}  & \frac{\partial F}{\partial v} \end{pmatrix} (\textbf{x})$$
i.e showing $\textbf{det}(T) \not = 0$ implies $F$ is a local diffeomorphism and now get your result. 
