Find the asymptotes of $xy^2 - y^2 - x^3 = 0$ I am lost with this question because there is no 3rd degree $y$ term and no 2nd degree $x$ term.
Can $y - x - 1/2 = 0$ and $y + x + 1/2 = 0$ be its asymptotes?
or $y - x - 1/3 = 0$ and $y + x + 1/2 = 0$
 A: Now, $y=\sqrt{\frac{x^3}{x-1}}$. Apart from the vertical asymptote $x=1$, we have two oblique asymptotes (OAs) in the form $y=ax+b$. To find these let us make the following limit zero:
$\begin{align}
\lim_{x\rightarrow \pm\infty}\sqrt{\frac{x^3}{x-1}}-(ax+b)&=\lim_{x\rightarrow \pm\infty}\frac{\frac{x^3}{x-1}-(ax+b)^2}{\sqrt{\frac{x^3}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{x^2+x+1+\frac{1}{x-1}-a^2x^2-2abx-b^2}{\sqrt{x^2+x+1+\frac{1}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}+\frac{1}{x-1}}+ax+b}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{|x+\frac{1}{2}|+ax+b}\\
\end{align}$
To make this limit zero at $\infty$: $a=1$ and $2b-1=0$. So, the OA at $\infty$ is $y=x+\frac{1}{2}.$
To make this limit zero at $-\infty$: $a=-1$ and $2b+1=0$. So, the OA at $-\infty$ is $y=-x-\frac{1}{2}.$
A: A general advice : in a preliminary step, always plot the curve(s) you have to study. It is so easy with free software like GeoGebra or Desmos !

The given implicit equation is equivalent to the two cartesian equations :
$$\color{red}{y=f(x)=x\sqrt{\frac{x}{x-1}}} \ \text{and} \ \color{blue}{y=g(x)=-x\sqrt{\frac{x}{x-1}}} \tag{1}$$
(blue and red curve resp.) Please note that both curves are disconnected with two branches : they are undefined for $x \in (0,1]$.
Let us work on the first function $f$.

*

*when $x\to 1_+$, $y \to +\infty$ ; therefore $x=1$ is a vertical asymptote.


*dealing with the right branch of this red curve, when $x \to \infty$, one can expand (generalized binomial expansion):
$$f(x)=x\sqrt{\frac{x-1+1}{x-1}}=x\sqrt{1+\underbrace{\frac{1}{x-1}}_u}=x(1+\frac12 u -\frac14 u^2+\frac38 u^3+\cdots)$$
$$f(x)=x+\frac12 \frac{x}{x-1}+...$$
(where the dots represent an expression that tends to 0 when $x \to \infty$).
Therefore, subtracting $\frac12$ to both sides of (1), we obtain :
$$f(x)-(x+\frac12)=\underbrace{\frac12(\frac{x}{x-1}-1)}_T+...$$
The first term $T$ in the RHS tends as well to $0$ when $x \to +\infty$ or $x \to -\infty$.
As the "gap" between $y=f(x)$ and
$$y=x+\frac12\tag{2}$$
tends to $0$, the line with equation (2) is the oblique asymptote to the red curve either for the right or the left branch.
Of course curve with equation $y=g(x)$ which is symmetrical to $y=f(x)$ wrt $x$-axis has as well 2 asymptotes which are the symmetrical straight lines wrt $x$-axis with resp. equations
$$ x=1, \ \ y=-x-\frac12$$
