If $F(x+y+z, x^2+y^2+z^2)=0$ then find $\frac{\partial^2 z}{\partial x \partial y }$ If $F(x+y+z, x^2+y^2+z^2)=0$ then find $\frac{\partial^2 z}{\partial x \partial y }$.
Attempt
I think that here we must apply the implicit differentiation Theorem, but I´dont know how I should do it, first I try use $X=x+y+z$ and $Y=x^2+y^2+z^2$ and then In case of be useful calculate
$$\frac{\partial X}{\partial x}=1, \, \frac{\partial X}{\partial y}=1, \, \frac{\partial X}{\partial z}=1$$ and also $$ \frac{\partial Y}{\partial x}=2x, \, \frac{\partial Y}{\partial y}=2y, \, \frac{\partial Y}{\partial z}=2z$$ and hence my Function should looks as $$F(X,Y)=0$$ From here I Try apply the implicit function theorem which states that I should find a function $z(X)$ such that $F(X,Z(X))=0$ and that in fact $z$ is differentiable with differential equal to $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$
But I´m not sure about if the form of I use actually is valid, and in other case someone can clarify what is the answer(step by step (because i´m learning Analysis by myself)  and more important how I should apply and understand this famous theorem.
 A: Apparently, it is required to express $\dfrac{\partial^2 z}{\partial x \partial y }
$ in terms of the derivatives $\dfrac{\partial F}{\partial X}$, $\dfrac{\partial F}{\partial Y}$, $\dfrac{\partial^2 F}{\partial X^2}$ etc. We will assume that all derivatives under consideration are continuous, so that the order of differentiation does not matter.
According to the implicit function theorem
$$
\frac{\partial z}{\partial x}=-\left.\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial x}\right/\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial z}.
$$
Using the chain rule, one obtains
$$
\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial x}=\frac{\partial F}{\partial X}
\frac{\partial X}{\partial x}+\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial x}=\frac{\partial F}{\partial X}+2x\frac{\partial F}{\partial Y}
$$
$$
\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial z}=\frac{\partial F}{\partial X}
\frac{\partial X}{\partial z}+\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial z}=\frac{\partial F}{\partial X}+2z\frac{\partial F}{\partial Y},
$$
(hereinafter, we mean by $\frac{\partial F}{\partial X}$
$\frac{\partial F}{\partial X}(x+y+z, x^2+y^2+z^2)$ etc.)
so
$$
\frac{\partial z}{\partial x}=-\frac{\dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}}{\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}}.
$$
Let us differentiate this with respect to $y$. According to the quotient rule
$$\tag{1}
\frac{\partial^2 z}{\partial x\partial y} =-
\frac{  
\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)-
\left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)
}
{
\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)^2}.
$$
Let's calculate $\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}\right)$ and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial Y}\right)$ using the chain rule:
$$
\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}\right)=
\dfrac{\partial}{\partial X}\left( \dfrac{\partial F}{\partial X}\right)
\frac{\partial X}{\partial y}+
\dfrac{\partial}{\partial Y}\left( \dfrac{\partial F}{\partial X}\right)
\frac{\partial Y}{\partial y}=
\dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X}
$$
$$
\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial Y}\right)=
\dfrac{\partial}{\partial X}\left( \dfrac{\partial F}{\partial Y}\right)
\frac{\partial X}{\partial y}+
\dfrac{\partial}{\partial Y}\left( \dfrac{\partial F}{\partial Y}\right)
\frac{\partial Y}{\partial y}=
\dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2}.
$$
This implies that
$$
\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)=
\dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X}+
2x\left(
\dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2}
\right)
$$
and
$$
\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)=\dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X}+
2\dfrac{\partial z}{\partial y}\dfrac{\partial F}{\partial Y}
+
2z\left(
\dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2}
\right).
$$
It remains to calculate $\dfrac{\partial z}{\partial y}$ in the same way as before we calculated $\dfrac{\partial z}{\partial x}$, and substitute all this into (1).
