Prove the equation implicitly represents a function Unfortunately I got stuck solving the following exercise:

Prove that the equation $xe^z + ye^{-z} + z = 1$ implicitly defines z as a function of (x,y), in the set $A\times[0,\infty)$ where: $A = \{(x,y)\in[0,\infty)\times[0,\infty)|x+y\leq1\}$.

Now, from what I understand, I am to show that z has a single solution in $[0,\infty)$ for all
$(x,y)\in{A}$ (where A can be described as all the points inside a triangle the Cartesian plain that its vertices are the origin, (1,0) and (0,1)). Or, in other words, z is a function of (x,y) in the specific region of topic.
I tried solving the equation to no avail (that is extracting z). It occurred to me that it might be unnecessary to solve the equation
for the proof but I can't seem to find any other way.
Any hints or directions will be gratefully accepted.
Thanks :)

Idea
Prove that $\exists!z\in[0,\infty)$ that satisfies the equation $\forall(x,y)\in{A}$ and to show that I first want to show that
$\frac{d(xe^z + ye^{-z} + z)}{dz} \geq 0$. Now show $xe^z + ye^{-z} + z \leq 1$ for $z = 0$ and due to the fact that
$xe^z + ye^{-z} + z$ is continuous as a function of $z$ and $\frac{d(xe^z + ye^{-z} + z)}{dz} \xrightarrow[z \to \infty]{} \infty$ the equation must be satisfied for a single $z$.
Nothing is mentioned about partial derivatives in my book up to this question so I am not sure it's even possible to claim what I've claimed here but I think this is the general idea. Don't really have anyway to confirm the validity of the proof and it is challenging for me to formalize it but I'll do my best and post an update.
