Numerical solution of differential equation Show that the nonlinear oscillator $y" + f(y) =0$ is equivalent to the system
$y'= -z $,
$z'= f(y)$
and that the solutions of the system lie on the family of curves
$2F(y)+ z^2 = constant $
where $F_y= f(y)$. verify that if $f(y)=y$ the curves are circle.
=> 
nonlinear oscillator $y" + f(y) =0$
where 
$y'= -z $,
$z'= f(y)$
so that means
$z''+z =0$
for the solution of the system lie on the family of curves, i was thinking
$\frac{d}{dt}[2F(y(t))+z^2(t)]= 2F \frac{dy}{dt} + 2z \frac{dz}{dt}$   
$=-2Fz +2zf(y)$                                 
$=-2f(y)z+2zf(y)$
$\frac{d}{dt}[2F(y)+z^2]=0$
$2F(y)+ z^2 = constant $
if $f(y)=y$ , then the differential equation is  $y'' + y =0$, meaning that
$y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$.
are the rotate axes.
$pA^2+qAB+rB^2=1$
$p,q,r$ depends on $x$
choose $x$ such that $q=0$
$pA^2+rB^2=1$
what can i do after that?
can someone please check my first,second and last part of answer.

For the first part, you made a mistake.
You calculated $$\frac{dy}{dt} (F(y)^2 + z^2) = y\frac{dy}{dt} + z\frac{dz}{dt}$$
which is completely untrue.
Instead, try to calculate
$$\frac{d}{dt}\left[2F(y(t)) + z^2(t) \right].$$
If $f(y)=y$, then the differential equation is $y'' + y = 0$, meaning that $y=A\cos x + B\sin x$ and $z=-y'= -A\sin x + B\cos x$.
