I have a system of non-linear equation such that $$f(x,y,z; \alpha, \beta) = 0,$$ $$g(x,y,z; \alpha, \beta) = 0,$$ $$h(x,y,z; \alpha, \beta) = 0,$$ where $\alpha \in [0,1]$, and $\beta \in (0,\infty)$. Let's fix the parameter $\beta$ as $\bar{\beta}$, I know the solution of this system in closed form, i.e. $x(\alpha, \bar{\beta}), y(\alpha, \bar{\beta}), z(\alpha, \bar{\beta})$ are known explicitly when $\alpha = 0$ and $\alpha = 1$. On top of that, the solution under these cases are unique. Can we say anything about the existence and uniqueness of $x,y,z$ for the other values of $\alpha$?
The argument about uniqueness could be hard, there might be something for existence, at least numerically it seems it has solution.
Note: These three functions are continuous in all parameters, and are in polynomial forms.
No. I can give examples any case you want. Say I want a set of equations where $\alpha=0$ and $\alpha=1$ have unique solutions, and nothing in between has a solution. I can safely ignore $\beta$ for this. Then a simple example is $$x-y=0\\x-z=0\\x^2+(4\alpha-2)x+1=0$$ When $\alpha=0$ you have $x=y=z=1$ and for $\alpha=1$ you get $x=y=z=-1$. For any $\alpha \in (0,1)$, the discriminant of the quadratic is $4[(2\alpha-1)^2-1]$, and since $|2\alpha-1|\lt 1$, you have no solution.
You can have an example where there is unique solution everywhere: $$x+\alpha+1=0\\y-\alpha-2=0\\3z-2\alpha=0$$