Finding the stationary point with implicit differentiation

Question

I want to find the stationary points of the curve $y^3+3xy^2-x^3=3$

I differentiate to get $\frac{x^2-y^2}{y^2+2xy}$

and so $x^2-y^2=0$ but I have two unknowns and I'm not sure how to solve.

Answer 1

Yeah you're on the right track... you just gotta keep going. \begin{align} \frac{x^2-y^2}{y^2+2xy} = 0\ \\ \\ \implies y(y+2x) \neq 0\quad \text{and}\quad (x+y)(x-y) = 0\\ \\ \implies (y \neq 0\quad \text{and}\quad y \neq -2x)\quad \text{and}\quad (x=y\quad \text{or}\quad x = -y).\\ \end{align} $$$$ Now substitute $x = y$ back into the original equation $\ y^3+3xy^2-x^3=3\ $ to find the coordinate of the stationary point corresponding to when $x = y$.

Then do a similar back-substitution for when $\ x = -y\ $ to find the coordinate of the other stationary point.