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:)

I'd try to give another answer. Recall from wikipedia and the most common definition of asymptotic line, to find an asymptotic line of $f(x)$ as $x\to\infty$, we need to compute 
(1) $a=\lim_{x\to\infty}\frac{f(x)}{x}$ 
(2) $b=\lim_{x\to\infty}(f(x) - ax)$. 
Then the asymptotic line would be $y=ax+b$. 
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By implicit function theorem when $x$ is large, $y$ can be written as a function of $x$.  Let's begin with 
$x^3 + 3x^2 y - 4y^3 -x + y + 3 = 0.   \qquad(1)      $
Then 
$1+ 3\frac{y}{x} - 4(\frac{y}{x})^3 - \frac{1}{x^2} + \frac{y}{x}\frac{1}{x^2} + \frac{3}{x^3}=0$
if $\lim_{x\to\infty} \frac{y}{x}=:a$ exists, then it might be a solution of $1+3a-4a^3=0$.  Moreover, the little consequence $\lim_{x\to\infty} \frac{y-ax}{x}=0$ is required.  
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Then we may try $x^3 + 3x^2(y-ax+ax) -4(y-ax+ax)^3 - x + (y-ax+ax) + 3 = 0$. After simplifying, we obtain 
$3(y-ax) - 4(y-ax) (\frac{y-ax}{x})^2 - 12 a(\frac{y-ax}{x})^2 - 12( y-ax)a^2 - \frac{1}{x} + \frac{y-ax}{x}\frac{1}{x} + \frac{a}{x} + \frac{3}{x^2}=0$. 
Thus 
$(y-ax)[3 -4(\frac{y-ax}{x})^2 -12a^2] = 12a(\frac{y-ax}{x})^2 + \frac{1}{x} -\frac{y-ax}{x}\frac{1}{x} - \frac{a}{x} -\frac{3}{x^2}$. Taking limit, we obtain $b(3-12a^2)=0$, where $b=\lim_{x\to\infty} y-ax$, existing. 
