Use Implicit Function Theorem I have these two equations $u + e^u + z +e^z = x$ and $u + u^5 + z^3 = y$. I'm wanting to try to show that there exist solutions $z = f(x,y)$ and $u = g(x,y)$ for $(x,y)$ in a neighborhood of $(2,0)$ and that $f$ and $g$ are smooth.I'm trying to figure this out and, in turn, figure out the bottom part as well. What i've tried so far is taking $$F_1(u,x,y,z)=u+e^u +z+e^z-x$$ $$F_2(u,x,y,z)=u+u^5+z^3-y$$
I've tried taking the determinate $$det=
        \begin{matrix}
        \frac {\partial F_1}{\partial u} & \frac {\partial F_1}{\partial z}\\
        \frac {\partial F_2}{\partial u} & \frac {\partial F_2}{\partial z} \\
        \end{matrix}
$$
however, I believe i'm approaching this the wrong way because I get$$
        \begin{matrix}
        1+ ue^u & 1+ze^z \\
        1+5u^4& 3z \\
        \end{matrix}
$$
I don't know where to plug in any other information and it seems I've run into a dead end.
I need to know this in order to do the example problems like:
Find $f(2,0)$ , $\frac {\partial f}{\partial x} (2,0)$ , $\frac {\partial f}{\partial y} (2,0)$ , $\frac {\partial^2 f}{\partial x^2} (2,0)$ , $\frac {\partial^2 f}{\partial y^2} (2,0)$ $\frac {\partial^2 f}{\partial x^2y} (2,0)$ Am I crazy or have i just screwed up big time somewhere?!?
 A: As mentioned in the comments, your question needs a little more info, because there are multiple points $(u,z)$ for which $x=2$, $y=0$.  However, notice that $(u,z) = (0,0)$ works, and I think that this is a natural point to choose.  So let's assume that the point $u=0$, $z=0$, $x=2$, $y=0$ was given in the problem instead of just $x$ and $y$.
Now if you weren't asked to verify that the implicit function theorem holds, your problem is much easier.  Let's rewrite your first equations, assuming that $u = u(x,y)$, $z=z(x,y)$ are functions of $x$ and $y$:
\begin{align*}
u(x,y) + e^{u(x,y)} + z(x,y) + e^{z(x,y)} & = x \\
u(x,y) + u(x,y)^5 + z(x,y)^3 & = y.
\end{align*}
(You don't always need to write all the dependencies on $(x,y)$, but it's nice to make sure you know exactly which variables are functions of which other variables.)
Now we can implicitly differentiate both of these with respect to $x$:
\begin{align*}
\frac{\partial u}{\partial x}(x,y) + e^{u(x,y)}\frac{\partial u}{\partial x}(x,y) +
\frac{\partial z}{\partial x}(x,y) + e^{z(x,y)}\frac{\partial z}{\partial x}(x,y) & = 1 \\
\frac{\partial u}{\partial x}(x,y) + 5u(x,y)^4\frac{\partial u}{\partial x}(x,y) + 3z(x,y)^2\frac{\partial z}{\partial x}(x,y) & = 0.
\end{align*}
Now plugging in $x=2$, $y=0$, $u(2,0)=0$, $z(2,0)=0$ gives:
\begin{align*}
2\frac{\partial u}{\partial x}(2,0) + 2\frac{\partial z}{\partial x}(2,0) & = 1 \\
\frac{\partial u}{\partial x}(2,0) & = 0,
\end{align*}
so $\frac{\partial z}{\partial x}(2,0) = \frac{\partial f}{\partial x}(2,0) = \frac12$.
You can do the higher derivatives similarly, though it looks like it will be quite messy.

By the way, since you ask about implicit function theorem, and you were close to showing that it holds, let's finish that off.  You compute the matrix of partial derivatives as you did above (but I've fixed some of yours which were incorrect):
$$\left( \begin{array}{cc} 1 + e^u & 1 + e^z \\ 1+5u^4 & 3z^2 \end{array} \right).$$
The determinant of this matrix is $3z^2(1+e^u) - (1+e^z)(1+5u^4)$.  Now we can plug in $u=0$, $z=0$, and notice that the derivative is nonzero.  By the implicit function theorem, this guarantees the existence of the functions $z=f(x,y)$, $u=g(x,y)$ which we assumed above.
