$$ y: (0, \sqrt{2}) \to \mathbb{R}, x \mapsto y(x) $$
And I have a function
$$ F(x,y) = (x^2+y^2)^2 - 2(x^2-y^2) $$
for which
$$ F(x, y(x)) = (x^2 + [y(x)]^2)^2 - 2(x^2 - [y(x)]^2) = 0 $$
How can I find the critical points of $y(x)$ or in other words, how can I find $y'(x)$ here? So far I got this: $$ \frac{\partial F}{\partial x}(x, y(x)) = 2(x^2 + [y(x)]^2) \cdot (2x + 2y(x) \cdot y'(x)) - 4x + 4y(x) \cdot y'(x) $$
$$y'(x)\left[4y(x)+2y(x)2(x^2+y^2(x)\right]+2\left[x^2+y^2(x)\right]2x-4(x)=\frac{\partial F}{\partial x}\left[x, y(x)\right]$$ $$ y'(x)= \frac{\left[\frac{\partial F}{\partial x}\left[x, y(x)\right]\right]-\left[2\left(x^2+y^2(x)\right)2x-4(x)\right]}{4y(x)+2y(x)2\left[x^2+y^2(x)\right]}$$ $$y'(x)=\frac{\left[\frac{\partial F}{\partial x}\left[x, y(x)\right]\right]-4x^3-4xy^2(x)+4x}{4y(x)+4x^2y(x)+4y^3(x)} $$