Prove that $\lim_{(x,y)\to (0.0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}$ does not exist I'm struggling with this limit. I have to approach using different curves and show that there is one curve wich prove that this limit does not exist, despite the fact that when trying with a lot of curves shows that the limit is 0.
If someone helps me to find that curve I will be very pleased. Thanks!
$$\lim_{(x,y)\to (0.0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}$$

By writing $x,y$ in polar coordinates,
$$\lim_{(x,y)\to (0,0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}=\lim_{r\to 0}\ r\cdot\frac{4\cos\theta\sin^2\theta - 3\cos^3\theta}{\cos^2\theta - \sin^2\theta}$$
indicating that the limit may not be zero only when $\tan^2\theta\to 1$, that is $\theta\to \frac{\pi}{4}$ or $\theta\to\frac{7\pi}{4}$.
Therefore, we shall choose a curve which the slope at origin is $\tan\frac{\pi}{4}=1$ or $\tan\frac{7\pi}{4}=-1$. For example, $y=x^2+x$ would be a choice.
$$\lim_{x\to 0} \frac{4x(x^2+x)^2 - 3x^3}{x^2 - (x^2+x)^2}=\lim_{x\to 0}\frac{4x^5+8x^4+4x^3-3x^3}{x^2 - x^4-2x^3-x^2}=\lim_{x\to 0}\frac{4x^2+8x+1}{-x-2}=-\frac{1}{2}$$
