Determining what set of points a curve can be expressed as a singlevariable-function The curve $$x^2y^3-3xy^2-9y+9=0$$ is given. I want to determine what points on the curve, for a neighbourhood to said points, $y$ can safely be expressed as a function of $x$. I guess what this means is that I need to find sets $A$ such that for $x_i \in A$ yields one and only one corresponding y-term $y_i$. Furthermore, this means that the y-term in the gradient in every such point is separate from zero, i.e. $$\frac{\partial f}{\partial y} (x_i, y_i) \neq 0$$ where $f$ is the the function $f : x \rightarrow y$. Would this be accurate? 
 A: An elementary way: assume $(x,y_1)$ and $(x,y_2)$ lie on the curve. Then
$$x^2y_1^3-3xy_1^2-9y_1+9=x^2y_2^3-3xy_2^2-9y_2+9$$
$$0=x(y_1^3-y_2^3)-3x(y_1^2-y_2^2)-9(y_1-y_2)$$
$$0=x(y_1^2+y_1y_2+y_2^2)-3x(y_1+y_2)+9$$
Now find when a real solution $y_2$ exists by taking a discriminant. As I'm working it out, it appears that we need to use the quadratic formula to solve for $x$ in terms of $y_1$, so it's not at all a pleasant bash...
Another approach - that for which you were probably looking - is finding the $x$-extrema by finding the critical points of the function.
A: I think I will post this picture, I think students here undervalue the skill of drawing; in this case, I am putting a computer picture, but it can be substantiated. The calculation part, for you to finish, is to correctly identify the two points where the graph is perfectly vertical, one in the first quadrant, one in the second. Also verify that the apparent asymptotes make sense, and so on. If you make no mistakes, the two vertical points come out with nice coordinates. If, as I did the first time, you make any small error, it comes out garbage and confusing. EDIT: for many of you, we are looking for the exact points where the gradient of the two-variable function is horizontal. This is the easiest case of Lagrange multipliers: to find the local extrema of the function $x$ constrained to the curve(s) $f(x,y) = x^2 y^3 -3x y^2 - 9 y + 9$ is set to $0,$ find the points on the curve where $\nabla f$ is a scalar multiple of $(1,0),$ which is to say, the points on the curve(s) where $$ \frac{\partial f}{ \partial y} = 0. $$ 

