How to find asymptotes of implicit function? How to find the asymptotes  of the implicit function $$8x^3+y^3-6xy-3=0?$$
 A: The asymptotes of an algebraic curve are simply the lines that are tangent to the curve at infinity, so let's go through that calculation.
First, we find where your curve meets the line at infinity. We homogenize to $(X:Y:Z)$ coordinates, so that $(x,y) = (X:Y:1)$. The equation is now
$$8X^3+Y^3−6XYZ−3Z^3=0$$
and the line at infinity is the line $Z=0$. Plugging that in and solving over the reals
$$ 8X^3 + Y^3 = 0
\\ Y = -2X$$
so $(1:-2:0)$ is the only (real) projective point where the curve meets infinity. (There will be two complex asymptotes as well, but we can't see them if we're only looking at geometry with real coordinates)
To find the tangent line, we note that the differential of the tangent line is the same as the differential the curve (more or less because the tangent lines have to point in the same direction). So we take the derivative:
$$ (24X^2 - 6YZ) dX + (3Y^2 - 6XZ) dY - (6XY + 9Z^2) dZ $$
and evaluate at $(1:-2:0)$ to get
$$ 24 dX + 12 dY + 12 dZ$$
and so the tangent line should be given by the equation
$$ 24 X + 12 Y + 12 Z = 0 $$
or, dehomogenizing:
$$ 24x + 12y + 12 = 0
\\ y = -1 - 2x $$
This is, of course, the same calculation as in the other answer, but if you're familiar with projective geometry (or are willing to self-study how it works), it explains how the calculation works.
A: The article you need to thoroughly put your eyes on is this Asymptote: Algebraic curves. So, let's do it step by step
1) Split your polynomial into homogeneous ones
\begin{align}
P_3(x,y) &= 8x^3+y^3 \\
P_2(x,y) &= -6xy \\
P_0 &= -3
\end{align}
$$
f(x,y) = 8x^3+y^3-6xy-3 = P_3(x,y) + P_2(x,y) + P_0 = 0
$$
2) Try to decompose $P_3(x,y) = Q(x,y)$ to the form of $(ax-by)Q_2(x,y)$, where $Q_2$ is another homogeneous polynomial. In this particular case it's quite easy to do, since
$$
P_3(x,y) = 8x^3+y^3 = (2x+y)\left( 4x^2-2xy+y^2\right )
$$
so $a = 2$ and $b = -1$.
3) Find the values of $Q_x'(b,a)$ and $Q_y'(b,a)$
\begin{align}
Q_x'(b,a) &= \left . 24x^2 \right |_{x = -1} = 24 \\
Q_y'(b,a) &= \left . 3y^2 \right |_{y = 2} = 12
\end{align}
so, they don't vanish simultaneously, and therefore
$$
Q_x'(b,a) x + Q_y'(b,a) y + P_2(b,a) = 24 x + 12 y + 12 = 0
$$
is an asymptote. Latter can be simplified as
$$
y = -2x - 1
$$
To make sure of it, you can draw that expression's plot.

PS: There are actually several plots, as a demonstration that asymptote doesn't depend on $P_0 = 3$, it can be $0$ or $100$.
A: I have seen you are interested in doing problems by Maple so, the following codes may help you machineary:
[> f:=8*x^3+y^3-6*x*y-3:
   t := solve(f = 0, y):
   m := floor(limit(t[1]/x, x = -infinity));

$$\color{blue}{m=-2}$$
[> h:=floor(limit(t[1]-m*x, x = -infinity));

$$\color{blue}{h=-1}$$
