jmc

Since we know that $\sqrt[3]{-8}=-2$ is a rational number, $$f(\sqrt[3]{-8})=|\lfloor -2\rfloor |=2.$$Continuing from here, we know that $-\pi$ is irrational, thus $$f(-\pi )=\lceil -\pi {\rceil }^2=(-3{)}^2=9.$$Because 50 is not a perfect square, $\sqrt{50}$ must be irrational as well, so $$f(\sqrt{50})=\lceil \sqrt{50}{\rceil }^2=8^2=64.$$Finally, we know that $\frac{9}{2}$ is a rational number, so $$f\left(\frac{9}{2}\right)=\left|\lfloor \frac{9}{2}\rfloor \right|=4.$$Therefore $$f(\sqrt[3]{-8})+f(-\pi )+f(\sqrt{50})+f\left(\frac{9}{2}\right)=2+9+64+4=\boxed{79}.$$