<s>[INST] 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:)
 [/INST] Giving 
$$
f(x,y) = x^3 + 3 x^2 y - 4 y^3 - x + y - 3 = 0
$$
the asymptotic directions can be explored by substituting $ y = a x + b\;\; $ into $f(x,y)$ giving
$$
f(x,ax+b) = 3 x^2 (a x+b)-4 (a x+b)^3+a x+b+x^3-x-3 = (1+3a-4a^3)x^3+3b(1-4a^2)x^2 +(a(1-12b^2)-1)x+b(1-4b^2)-3
$$
Now the conditions for  $f(x,ax+b)$ to have line behavior are 
$$
\left\{
\begin{array}{rcl}
 3 b-12 a^2 b=0 \\
 -4 a^3+3 a+1=0 \\
\end{array}
\right.
$$
and solving we find two solutions: 
$$
\{a =-\frac 12, \forall b\} \cup \{a = 1, b = 0\}
$$
so we have
$$
L_1\to y = x\\
L_2\to y = -\frac x2 + b
$$
Considering $f(x,y)$ at their asymptotic values it's value should be the same or
$$
\lim_{x\to 0}f(x,-\frac 12 x+b)=\lim_{x\to 0}f(x,x) = -3
$$
so this can be solved by choosing $b$ such that $1-4b^2 = 0\;\;$ giving
$$
L_1\to y = x\\
L_2\to y = -\frac x2 \pm\frac 12
$$
Attached a plot showing in blue $f(x,y) = 0$ and in red $L_1, L_2$

</s>