Evaluate derivatives $y'(0)$, $z'(0)$, $y''(0)$ ,$z''(0)$ of implicit functions $y(x)$ and $z(x)$, where $y(0)=-1$ and $z(0)=1$, given by system of equations: $x+y+z=0$ and $x^2+y^2+z^2=0$
First step: I evaluated $y(x)$ as $x^2+xy+y^2=0$
But I have no idea how $y(0)$ can be equal to $-1$.
By the implicit function theorem, you know that the functions $y$ and $z$ exist and are defined for $x\in(-\epsilon,\epsilon)$ for some small enough $\epsilon$. You also know that for these functions, the following things hold (for all values of $x\in(-\epsilon, \epsilon)$: $$x+y(x) +z(x) =0\\ x^2+(y(x))^2 + (z(x))^2 = 0\\ y(0)=-1\\ z(0)=1$$
Now, if I differentiate the first equation, I get that $$1 + y'(x) + z'(x) = 0,$$ and the second equation gives me $$2x + (y(x)^2)' + (z(x)^2)' = 0$$ for $x\in(-\epsilon, \epsilon)$. Of course, $(y(x)^2)'$ is simply $2y(x)y'(x)$. What happens if I take a look at the values at $x=0$? What equations do you get?