I'm currently working through the problem:
In a neighbourhood of $(u,v)=(1,1)$; show that the system:
$$ \begin{cases} x=u+\ln(v) \\ y = v-\ln(u) \\ z=2u+v \end{cases} $$
defines a function $z=z(x,y)$ and find the partial derivatives $z'_{x}$ and $z'_{y}$
Now I know I am supposed to use the implicit function theorem in some way, but I'm used to starting with a system of equations like:
Example: $$ \begin{cases} F(x,y,u,v) = x^2 -y^2-2u=0 \\ G(x,y,u,v) = xy -v-u=0 \end{cases} $$
My gut instinct is telling me to use some substitution to see if I can't tease out $z=z(x,y)$ but I don't think it is that simple.
I guess I'm looking for a solid first step (or two steps) to get me started on tackling a question like this.
Given the system, you can "manually unroll" the implicit function theorem.
You can always differentiate the first two equations withe respect to $x$ and $y$ - you will get $4$ equations for four variables $u'_x$, $u'_y$, $v'_x$, and $v'_y$. This will give the expressions for these derivatives in terms of $u$ and $v$.
After that, you will need to find values of $u$ and $v$ corresponding to $(x,y)=(1,1)$.
Finally, $z'_x = 2u'_x+v'_x$ and similarly for $z'_y$.