AP Calculus Derivative Slope $xy^2-x^3y = 2$
$y' = \frac{3x^2y-y^2}{2xy-x^3}$
find the x-coordinate of each point on the curve where the tangent line is horizontal.
I know that $3x^2y-y^2 = 0 \Rightarrow$ then $y=3x^2.$
If I plug y into the original I get 
$x(3x^2)^2-x^3(3x^2)=2 \Rightarrow x(9x^4)-3x^5 = 2 \Rightarrow 9x^5 - 3x^5 = 2 -> x^5 = \frac{1}{3} \Rightarrow x = (\frac{1}{3})^{(\frac{1}{5})}$
Is this all I have to do?

Yes, you've found the unique solution: $\quad x = \left(\frac 13\right)^{1/5}$. 
Note: It also happens to be true that at $y = 0$, then $y' = 0$. But it turns out that $y = 0$ does not satisfy the original equation, and so does not lie on the curve. Hence, we have only the solution that you found.
