How to select points from a table that are within some region

You should parametrize your region. c looks to me like sort of a radius and it will help to cut off half a plane.

tbl=Flatten[Table[{r,z},{r,0,1,.1},{z,-.8,.8,.1}],1];

r[η_,ξ_, c_]:=(c Sin[η])/(Cosh[ξ]-Cos[η]);
z[η_,ξ_, c_]:=(c Sinh[ξ])/(Cosh[ξ]-Cos[η]);

η0=Pi/3;
ρ=(2/(3 (π-η0) Cos[η0]+3 Sin[η0]-Sin[η0]^3))^(1/3);
c0=ρ Sin[η0];

rig = Region[ParametricRegion[{{r[η0, ξ, c], z[η0, ξ, c]}, 
    0 < c < c0}, {{ξ, -30, 30}, c}], Epilog -> Point[tbl]]

sel=Select[tbl,RegionMember[rig,#]&];

Region[ParametricRegion[{{r[η0,ξ,c],z[η0,ξ,c]},0<c<c0},
{{ξ,-30,30},c}],Epilog->Point[sel]]

I am not sure why some points look they slightly sticking out of the region, - might be drawing imprecision or an issue. You should check this as I am out of time, I hope it's a start.

As suggested i present an alternative solution.

But keep in mind, this depends on the ability to solve your implicit equations and therefore, does not work everytime. But when it works it yields a fast Boolean condition.

We doing the following. Consider your set of equations, depending on the parameter $\xi$.

$$(x,y)=(r(\xi),z(\xi))$$

We solve the first one for the parameter and plug this solution into the other equation. We hope to not only solve this, but to get at least 2 equations which bound your region in $y(x)$.

$$\xi=r^{-1}(x)\\ y(x)=z(r^{-1}(x))$$

We can get the $r^{-1}$ with

xiSols=Normal[Solve[x==r[\[Eta]0,xi],xi,Reals]]

which has two solutions. So we've already won. Plug this into the second equation:

{ySolL,ySolR}=z[\[Eta]0,xi]/.xiSols//Simplify

The sign signals us that ySolL < ySolR and therefore ySolL < y < ySolR

We can easily check this:

RegionPlot[ySolL<=y<=ySolR,{x,0,1.2},{y,-1,1}]

So you could now either:

• create a region and check with Element

• Use the fast Boolean condition directly ySolL<=y<=ySolR && x > 0

  tblIn=Select[tbl,(ySolL<=y<=ySolR&&x>=0)/.{x->#[[1]],y->#[[2]]}&];

  With the plot

  Show[RegionPlot[ySolL <= y <= ySolR, {x, 0, 1.2}, {y, -1, 1}], ListPlot[tblIn]]