Suppose I have three functions $f(x,y)$, $g(z)$ and $h(z)$. These three functions are differentiable and continuous in the reals.
I am interested in the intersection: $$ z^*=f(x^*,y^*)\\ x^*=g(z^*)\\y^*=h(z^*) $$
I know that $f(x,y)$ is monotonically increasing in x and y, and the cross-derivative is zero (i.e. $\frac{\partial^2 f(x,y)}{\partial y \partial x}=0$). I also know that both $g(z)$ and $h(z)$ are monotonically decreasing in z.
Is this sufficient to prove that the set of values $(x^*,y^*,z^*)$ exists and is unique?
Yes, those conditions are sufficient (the differentiability is redundant). Consider the function $s \colon \Bbb R \to \Bbb R$ defined by $s(z)=f(g(z),h(z))-z$. It is continuous and strictly decreasing since $f(g(z),h(z))$ is decreasing. It is clear that the equation $$s(z)=0 \quad (1)$$ is equivalent to $f(g(z),h(z))=z$ and any $z_0$ satisfying it gives the solution $(x^*, y^*, z^*)=(g(z_0), h(z_0), z_0)$ of the original problem. Conversely, any solution $(x^*, y^*, z^*)$ of the original problem gives the solution $z_0=z^*$ of $(1)$.
Now the explanation why $(1)$ has a solution. Let $a=s(0)=f(g(0),h(0))$. If $a=0$, then $z_0=0$ is the solution.
If $a>0$, then $s(0)=a>0$, $s(a)=f(g(a),h(a))-a \le f(g(0),h(0))-a=a-a=0$. By the intermediate value theorem there is a point $z_0 \in (0,a]$ such that $s(z_0)=0$.
If $a<0$, the reasoning is similar: $s(0)=a<0$, $s(a)=f(g(a),h(a))-a \ge f(g(0),h(0))-a=0$. By the intermediate value theorem there is a point $z_0 \in [a,0)$ such that $s(z_0)=0$.
$(1)$ cannot have more than one solution because $s$ is strictly decreasing. Thus, the solution found is unique.