You can find the stationary points of $z=z(x)$ by application of the implicit function theorem (to find the derivative of $z$) and solving (numerically) a system of two equations:
stationarypts = {x, z} /. NSolve[{2 Ωxz - C0 == 0, -D[2 Ωxz - C0,x]/D[2 Ωxz - C0, z] == 0}, {x, z}]{{-1.0898108637505286, 0. + 0.4183859620455661*I}, {-1.0898108637505286, 0. - 0.4183859620455661*I}, {1.1719884857449565, 0. + 0.07308908417888743*I}, {1.1719884857449565, 0. - 0.07308908417888743*I}, {0.9892920541322313, -0.0984137896912804}, {-0.01369302911651984, 0.6235986400369145}, {-0.01369302911651984, -0.6235986400369145}, {0.9892920541322313, 0.0984137896912804}, {0.8365517541212919, 0. + 0.011277418446297614*I}, {0.8365517541212919, 0. - 0.011277418446297614*I}}
For the range of values of $x$ you are interest in, there are two stationary points:
Cases[stationarypts, x_ /; xL1 <= x[[1]] <= xL2]
{{0.989292, -0.0984138}, {0.989292, 0.0984138}}
ContourPlot[2 Ωxz - C0 == 0, {x, xL1, xL2}, {z, -0.2, 0.2}, Epilog -> {PointSize[Large], Point[Cases[stationarypts, x_ /; xL1 <= x[[1]] <= xL2]]}]
