Show that the line x+y=0 is an asymptote of the curve

Show that the line $x+y=0$ is an asymptote of the curve $xy^2+x^2y-8=0$.

Step 1: Factor

\[ xy^2+x^2y = xy(x+y) \;\Longrightarrow\; xy(x+y) = 8. \tag{1} \]

Step 2: Behaviour along the branch $y\approx-x$

From (1), directly, \[x+y = \frac{8}{xy}. \tag{2}\]

Along the branch where $y\approx-x$ as $(x,y)\to\infty$, $xy\approx-x^2\to-\infty$, so $x+y=8/(xy)\to0$.

Step 3: Cross-check via the homogeneous-part method

The leading (degree-3) part of $F(x,y)=xy(x+y)-8$ is $\Phi_3(x,y)=x^2y+xy^2$; setting $y=kx$, $\Phi_3(1,k)=k+k^2=k(1+k)=0$ gives asymptotic directions $k=0$ and $k=-1$. For $k=-1$ (slope $-1$), since there is no degree-2 part ($\Phi_2\equiv0$), the intercept $c$ solves $c\,\Phi_3'(1,k)+\Phi_2(1,k)=0$ with $\Phi_3'(1,-1)=1+2k|_{k=-1}=-1$, giving $c\cdot(-1)+0=0$, i.e.\ $c=0$.

Step 4: Verification

\[\text{Solving }xy(x+y)=8\text{ numerically for large }x\text{ (}x=50,200,1000\text{) gives }x+y\approx1.6\times10^{-3},\,1.0\times10^{-4},\,4.0\times10^{-6}\to0\ \checkmark\]