$\sqrt{2-x^2}+\sqrt[3]{3-x^3}=0$=$( \sqrt{\sqrt{2}-x})(\sqrt{\sqrt{2}+x})+\sqrt[3]{3-x^3}=0$
If $x>\sqrt{2}$ then obviously no solution would exist since $( {\sqrt{2}-x})<0$.
If $x<\sqrt{2}$ then $( \sqrt{\sqrt{2}-x})(\sqrt{\sqrt{2}+x})+\sqrt[3]{3-x^3}>0$.
So only remains to check $x=\sqrt{2}$ which doesn't work.