I am trying to do this exercise on Real Analysis topics. It's about implicit function and I don't know how to conclude anything. Here it goes:
"Consider the equation $x^3 + y^3 - 3axy = 1$ where $a > 0$. According to the implicit function theorem, characterize the set $D$ of the points $(x,y)\in R^2$ where the equation defines a function $y=y(x)$. Then, calculate the point $(x,y)\in D$ which satisfies $y'(x) = 0$."
I tried to think of three cases: the points $(x,0)$, $(0,y)$ and $(x,y)$ and I got to the condition $y^2\neq ax$. I thought this condition could define $D$, but then I found out the expression for $y'(x)$ and can't find its zero.
I'd be very grateful if anyone could help me answer the question. Many thanks!
we get for $y=y(x)$ the first derivative as follows: $$3x^2+3y^2y'-3ay-3axy'=0$$ solving for $y'$ we obtain: $$(3y^2-3ax)y'=3ay-3x^2$$ and finally $$y'=\frac{ay-x^2}{y^2-ax}$$ if $y^2-ax\neq 0$ for $y'(x)=0$ you get $ay-x^2=0$ and this can you set in your equation $$x^3+y^3-3axy=1$$