<s>[INST] Is there a special case for equations of curves that have parallel oblique asymtotes? I am given the function $f(x,y)=(x-y)^2(x^2+y^2)-10(x-y)x^2+12y^2+2x+y$  and asked whether the curve $f(x,y)=0$ has two parallel oblique asymptotes, two unique asymptotes, or if there is no oblique asymptote at all. Do I manually find the asymptote equations or there is some shortcut to this?
 [/INST] The algebraic approach here is not too easy, so here's how to blend in some calculus (and some nontrivial computation, perhaps done with the help of a CAS). Of course, using a graphical CAS, you can easily see that there are two asymptotes with slope $1$. My original guess was that $y=x$ was an asymptote because of the leading factors of $x-y$. I was wrong.
Let $f(x,y) = (x-y)^2(x^2+y^2)-10(x-y)x^2+12y^2+2x+y$.
Using implicit differentiation, we can examine the limiting slope of the level curve $f(x,y)=0$ when $y=mx+b$ as $x\to\infty$. The leading behavior is
$$\frac{dy}{dx} \approx \frac{m^3-2m^2+3m-2}{2m^3-3m^2+2m-1} = \frac{(m^2-m+2)(m-1)}{(2m^2-m+1)(m-1)},$$
and this has limit $m$ precisely when $m=1$. So we look for slant asymptotes $y=x+b$.
Let's review what it means for $y=x+b$ to be an asymptote of the curve $y=g(x)$. This should mean that the vertical error between the curve and the line approaches $0$, i.e., 
$$\lim_{x\to\infty} \big(g(x)-(x+b)\big) = 0.$$
Now, consider our level curve $f(x,y)=0$. Using linear approximation, we have 
$$0 = f(x,y) = f(x,x+b) + \frac{\partial f}{\partial y}(x,x+b)\big(y-(x+b)\big),$$
and so our vertical error is given by
$$y-(x+b) = -\frac{f(x,x+b)}{\frac{\partial f}{\partial y}(x,x+b)}. \tag{$\star$}$$
Start plugging away, substituting $y=x+b$ (so $x-y=-b$):
\begin{align*}
f(x,x+b) &= b^2(x^2+(x+b)^2)+10bx^2+12(x+b)^2 + 2x+x+b \\ &= 2(b^2+5b+6)x^2+2(b^3+12b)x+\dots.
\end{align*}
This will be smallest (for large $x$) when the leading coefficient vanishes, i.e., when $b=-2$ or $b=-3$. In these cases, $f(x,x+b)$ will be given by a linear polynomial. On the other hand, considering ($\star$), 
$$\frac{\partial f}{\partial y}(x,x+b) = 2(2b+5)x^2 + \dots$$
is quadratic for both $b=-2$ and $b=-3$, so the quotient giving $y-(x+b)$ is the quotient of a linear polynomial by a quadratic polynomial, and hence goes to $0$ as $x\to\infty$.
Whew! In sum, the asymptotes are $y=x-2$ and $y=x-3$, as, indeed, @quasi commented.
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