Two variables limit question I proved that $f(x,y)= \dfrac{xy^2}{x^2 + y^3}$ does not have limit at origin.
I used two paths test; first I followed the $x$ axis, then I followed $x = \frac{1}{2}(y^2 + (y^4 - 4y^3)^{1/2})$ for $y&lt;0$. 

However, I am STILL looking for other solutions other ideas.

Any kind of answer, help or hint is appreciated.
Think of the path $(x,-x^{2/3})$ in the open fourth quadrant as $x\to 0^+.$ The denominator of your expression equals $0$ at every point of this path.  Meanwhile the numerator is never $0$ in that open quadrant. That should give you pause. Even if you exclude this path from the domain of $f$ (which you should have done, otherwise you are dividing by $0$), you have insurmountable problems. There is no way around it: $f$ is unbounded in every neighborhood of $(0,0),$ so there is no hope of a limit.
Use polar coordinates, so your limit becomes

$$\lim_{r\to0}\frac{r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta+r^3\sin^3\theta}=\lim_{r\to0}\;r\frac{\cos\theta\sin^2\theta}{\cos^2\theta+r\sin^3\theta}=0$$