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?

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.
