I'm looking for a continuously differentiable parametrization of $$x^3+y^2-z^2=1$$ but I'm actually totally stuck. If the $x$ term were quadratic instead of cubic, it would be simple: $$(x,y,z)=(\sqrt{t^2+1}\cos\theta, \sqrt{t^2+1}\sin\theta, t)$$ But with the cubic term there, I'm stuck. I naturally thought about $$(x,y,z)=(\sqrt[3]{t^2+1}\cos^{\frac{2}{3}}\theta, \sqrt{t^2+1}\sin\theta, t)$$ but this isn't continuously differentiable in $\theta$.
Hints or suggestions?
