Check if $f(u,v)=(u+u^2,v^2,v^3)$ defines a regular surface for $(u,v)\in\mathbb{R}^2$.
If it was a regular surface then the jacobian of the $f$ would have rank $2$, but for $u=-\frac{1}{2}$ it fails. (Morever $f$ would need to be injective but $f(-1,0,0)=f(0,0,0).$)
The points of $f$ satisfy $x\geq-\frac{1}{4}$ and $z=y^2$. The points where $x=-\frac{1}{4}$ are the ones that fail the previous condition. I think the image of $f$ is not regular surface because the tangent plane is not defined on those points, but I do not know how to justify it rigorously.