How to solve this equation for oblique asymptotes $$x^3 +2x^2y -xy^2 - 2 y^3+ 3xy +3y^2+x+1=0$$
I do not understand the method described in my college book. It involves finding roots of highest degree terms and further uses the formula 
$$c= \frac{-Ф_{n-1} (m)}{Ф’_n (m)}$$
$Ф_n(m)$ is obtained by $x=1$, $y=m$ in highest degree terms of given equation
$Ф’_n$ is derivative of $Ф_n$.
general eq. used is $y=mx+c$
I don't get it. I can upload a picture if its allowed.
 A: They deliberately made this so it all would come out nice. The highest degree part is
$$ x^3 + 2 x^2 y - x y^2 - 2 y^3 = (x + 2 y)(x+y)(x-y) $$
so those give you the three values for $m,$ as $y \approx -x/2$ or  $y \approx -x$ or  $y \approx x.$ In all three cases, we then add a constant term called $c$ so that the highest degree terms remaining cancel, and the error in saying $y \approx mx + c$ goes to $0$ as $x \rightarrow \infty.$ I did that by hand, no idea what is in your book, here is a combined graph. The three oblique asymptotes are
$$ y = -x, \; \; \;  y = x + 1, \; \; y = -\frac{x}{2} + \frac{1}{2} $$ 
 
The reason that the $x + y$ factor gets constant $0$ is that $x+y$ also is a factor of the next lower degree terms, that is 
$$ 3xy + 3 y^2 = 3 y (x+y). $$
How to check this? Write one of them with a symbol for error, say
$$ y = -x + \varepsilon, $$ substitue that value for $y$ into the original equation, then solve for $\varepsilon$ as a function of $x.$ If I am correct, then you will find that $\varepsilon \rightarrow 0$ as $|x| \rightarrow \infty.$ Same for the other two, 
$$ y = x + 1 + \varepsilon, $$ finally
 $$ y = -\frac{x}{2} + \frac{1}{2}  + \varepsilon. $$
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