Csdrhrt

First of all, I'm sorry for any English mistakes I might make, it's not my first language.
I have some trouble understanding how to apply the implicit function theorem when there are several equations involved (don't know if I'm being entirely clear).

It is fairly straightforward when you have a function, say $F(x,y,z)=0$ and you want $z$ as a function of $x$ and $y$, but I've come across some exercises where there is more than one equation (or "function"?) involved.
I think it will be more clear if I show you an example:

"The three equations

$x^{2}-ycos (uv)+z^{2}=0$,

$ x^{2}+y^{2}-sin (uv)+2z^{2}=2$, and

$xy-sin ucos v+z=0$

define $x$ ,$y$, and $z$ as functions of $u$ and $v$. Find the partial derivatives $frac{partial x}{partial u}$ and $frac{partial x}{partial v}$ at the point $x=y=1, u=frac{pi }{2}, v=0, z=0$.

How would you apply the implicit function theorem here?

It's exactly the same as your example with the equation $F(x,y,z)=0$. The difference is that you have to change the number of dimensions.

Let $F:{ mathbb{R} }^{ 5 }rightarrow { mathbb{R} }^{ 3 }$ be:
$$
F(x,y,z,u,v)=({x}^{2}-ycos {(uv)+{ z }^{ 2 },quad{ x }^{ 2 }+{ y }^{ 2 }-sin { (uv) } +2{ z }^{ 2 }-2,quad xy-sin { (u)cos { (v) } +z) } }
$$

So now, your equation is $F(x,y,z,u,v)=(0,0,0)$.

Then we have to prove that we could put $x$, $y$ and $z$ as functions of $u$ and $v$, which is the same as saying $exists G(u,v)=(x,y,z)$. This is done using the implicit function theorem, that is composed of three steps:

In the first step we have to see if $F$ is a function of first class. In fact, $Fin { C }^{ infty }({ mathbb{R} }^{ 5 })$ so it is obviously that it satisfies the first condition.

The second step is to see if the point that we are studying is agree with the equation. True again, because $F(1,1,0,frac { pi }{ 2 } ,0)=(0,0,0)$.

Finally, we have to see if this is true at the given point:
$$
detleft( frac { partial F }{ partial (x,y,z) } right) neq 0
$$

We know $F$, so we only have to make some operations:
$$
F'(x,y,z,u,v)=begin{pmatrix} frac { partial F }{ partial (x,y,z) } & frac { partial F }{ partial (u,v) } end{pmatrix}
$$
$$
frac { partial F }{ partial (x,y,z) } =begin{pmatrix} 2x & -cos { (uv) } & 2z \ 2x & 2y & 2z \ y & x & 1 end{pmatrix}
$$
$$
detleft( frac { partial F }{ partial (x,y,z) } (1,1,0,frac {pi}{2},0) right) =2neq 0
$$

So, at the end, we can happily say that $exists G(u,v)=(x,y,z)$, thanks to the implicit function theorem.

Now you want to find the partial derivatives of $x(u,v)$. So, we apply the implicit derivative:
$$
G'(frac { pi }{ 2 } ,0)=-{ left( frac { partial F }{ partial (x,y,z) } (1,1,0,frac { pi }{ 2 } ,0) right) }^{ -1 }left( frac { partial F }{ partial (u,v) } (1,1,0,frac { pi }{ 2 },0) right)
$$
$$
G'(frac { pi }{ 2 } ,0)=-{ begin{pmatrix} 2 & 1 & 0 \ 2 & 2 & 0 \ 1 & 1 & 1 end{pmatrix} }^{ -1 }begin{pmatrix} 0 & 0 \ 0 & frac { -pi }{ 2 } \ 0 & 0 end{pmatrix}=begin{pmatrix} 0 & frac { pi }{ 2 } \ 0 & -pi \ 0 & frac { pi }{ 2 } end{pmatrix}
$$
$$
G'(frac { pi }{ 2 } ,0)=begin{pmatrix} nabla x(frac { pi }{ 2 } ,0) \ nabla y(frac { pi }{ 2 } ,0) \ nabla z(frac { pi }{ 2 } ,0) end{pmatrix}Rightarrow begin{cases} frac { partial x }{ partial u } (frac { pi }{ 2 } ,0)=0 \ frac { partial x }{ partial v } (frac { pi }{ 2 } ,0)=frac { pi }{ 2 } end{cases}
$$

I hope that this can help you. Sorry for my English skills and let me know if there is any error on my explanation.