Let F[y] be a functional defined like so: F[y] = $\int y(x)^2 + (y'(x))^2 dx$.
I'm trying to find the function y which maximizes the value of F, and because the Euler Lagrange equation specifies a necessary condition for an extremal y, this looks like the place to start: $$\frac{\partial{[y(x)^2 + (y'(x))^2]}}{\partial{y(x)}} - \frac{d}{dx} [ \frac{\partial{[y(x)^2 + (y'(x))^2]}}{\partial{y'(x)}}] = 0$$
In the process of producing a differential equation where I can solve for y(x), I apply this transformation:
$$ 2 y(x) + \frac{\partial{[y'(x)^2]}}{\partial{y(x)}} - 2 y''(x) - \frac{d}{dx} [\frac{\partial{[y(x)^2]}}{\partial{y'(x)}}] = 0$$
Now I'm stuck. What does $ \frac{\partial{[y'(x)^2]}}{\partial{y(x)}} $ mean? What does $ \frac{d}{dx} [\frac{\partial{[y(x)^2]}}{\partial{y'(x)}}] $ mean? What theorems/rules allow me to simplify those expressions? Or am I going down the wrong path?
When applying the Euler-Lagrange equations, you treat $y$ and $y'$ as separate entities, and therefore, both of the derivatives about which you inquire are zero. However, there is a mistake in your evaluation. It should read
$$2y(x)+\frac{\partial [(y'(x))^2]}{\partial y}-2y''(x)-\frac{d}{dx}\frac{\partial [y(x)^2]}{dy'}=0.$$
You forgot to apply the $x$ derivative to $y'$.