Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
 $$ 2y + x = ±1 $$ can't be deduced.
Instead another erroneous line $ y = 0 $ is outputted.
Some help will be appreciated.Alternate methods of solution would work too:)

If the slant asymptote exists, we may denote
$$k=\lim_{x \to \infty}\dfrac{y}{x}.$$
Then $$0=\lim_{x \to \infty}\left(1+3\cdot \frac{y}{x}-4 \cdot \frac{y^3}{x^3}-\frac{1}{x^2}+\frac{y}{x^3}+\frac{3}{x^3}\right)=1+3k-4k^3=-(k-1)(2k+1)^2.$$
Thus,$$k=1,-\frac{1}{2}.$$



*

*If $k=1$, we denote $y=x+b$ and put it into the equation, we have $$-4b^3-12b^2x-9bx^2+b+3=0.$$


Then $$0=\lim_{x \to \infty} \frac{-4b^3-12b^2x-9bx^2+b+3}{x^2}=-9b.$$
Thus,$$b=0.$$



*

*If $k=-\dfrac{1}{2}$, we denote $y=-\dfrac{x}{2}+b$ and put it into the equation, we have $$-8b^3+12b^2x+2b-3x+6=0.$$
Then $$0=\lim_{x \to \infty} \dfrac{-8b^3+12b^2x+2b-3x+6}{x}=12b^2-3.$$
Thus, $$b=\pm \frac{1}{2}.$$

As a result, the slant asymptotes are  $$y=x, y=-\frac{x}{2}\pm \frac{1}{2}.$$
