Oblique asymptotes of non rational, non function

Question

I'm given the equation: $6x^2y+xy^2-2x^3=0$ and I'd like to know the slope of the two lines it forms.

While graphing a more general form: $ax^2y+bxy^2-cx^3=k$, I've noticed that if you set the equation equal to something besides $0$, it resembles a rational in that it has two slant asymptotes of the equation set $=0$ and a vertical asymptote at $x=0$. Furthermore, increasing a rotates the asymptotes evenly clockwise and increasing b or c changes the angle between the asymptotes.

I also noticed, using implicit differentiation, that $\frac{dy}{dx}=\frac{6x^2-12xy-y^2}{6x^2+2yx}$ and by plugging into the equation of a line $y=\frac{dy}{dx}x$ gives $y=\frac{6x^2-12xy-y^2}{6x+2y}$ which perfectly matches the original graph of $6x^2y+xy^2-2x^3=0$.

I'm not sure how to approach this one since all the methods I know of for finding the equation of asymptotes only apply to functions.

Link to Desmos graph I used to graph the equation: https://www.desmos.com/calculator/jixvhuhmzt

Answer 1

for the first one, take $x \neq 0,$ divide through by $x^3.$ The original is homogeneous. Next, define $$ r = \frac{y}{x} $$ Your item becomes a polynomial in $r$ being set to zero. I get $$ 6r + r^2 - 2 = 0 $$

Answer 2

We have that

$$6x^2y+xy^2-2x^3=0 \iff xy^2+6x^2y-2x^3=0$$

$$y=\frac{-6x^2\pm\sqrt{36x^4+8x^4}}{2x}=-3x\pm x\sqrt{11}$$

Answer 3

$6x^2y+xy^2-2x^3=0\implies x=0 $ or $y^2+6xy-2x^2=0.$

$y^2+6xy-2x^2=0\implies (y+3x)^2=11x^2\implies y+3x=\pm\sqrt{11}x\implies y=(-3\pm\sqrt{11})x,$

so the slopes of the two lines you'd like to know are $-3\pm\sqrt{11}$