We are given the function $$ F(x,y) = (x^2 + y^2)^2 - 2(x^2 - y^2) $$
with the condition $F(x, y(x)) = 0$ for $$ y: (0, \sqrt{2}) \to \mathbb{R}, x \mapsto y(x) $$
The objective is to compute all extreme points and classify them into max/min. So far I've computed the partial derivatives and got an expression for $y'$ $$ \begin{align*} \partial_x F(x,y) &= 4x(x^2 + y^2 - 1) \\ \partial_y F(x,y) &= 4y(x^2 + y^2 + 1) \end{align*} $$
So using the chain rule we get $$ y' = - \frac{\partial_x F(x, y(x))}{\partial_y F(x,y(x))} = - \frac{x(x^2 + [y(x)]^2 - 1)}{y(x) (x^2 + [y(x)]^2 + 1)} $$
So looks like $y' = 0 \iff x = 0 \lor x^2 + [y(x)]^2 - 1 = 0$.
Now because of the constraints given on $y(x)$, $x \in (0, \sqrt{2}) \implies x \neq 0$.
Thus the only possible solution is $x^2 + [y(x)]^2 - 1 = 0$ which implies $x^2 + [y(x)]^2 + 1 = 2$ and $y(x) \neq 0$.
But how can I compute the exact $x$ value from this?
Using $$y^2=1-x^2$$ we get $$1=2(x^2-(1-x^2))$$