$x^3+y^3=\frac{1}{\sqrt{2}}$ is according to my textbook not bounded since, for each $x_0 \in R$ there is a $y_0$ such that the point $(x,y_0)$ is on the curve.
Isn't that true for a bounded set also e.g $x^2+y^2=1$. Im not sure what this statement says exactly.
Nope, that is not true, the difference between these 2 cases is that equation $y^3=a$ will have a real solution for any $a$, but $y^2=a$ will have a solution only for $a\geq0$, so if $|x|>1$ you will not find such $y_0$