Confused with the notation/halfway done and need a little help I am supposed to solve the following problem:
Given 1 and 2
1) $x^{2}+y^{2}+z^{2}=6$
2) $w^{3}+z^{3}=5xy+12$
I must solve for $\frac{\partial x}{\partial w}$ at a constant $z$ at the point (x,y,z,w) = (1,-2,1,1) and the answer is -6/5, but I can't seem to get it.
Taking the partial derivative with respect to x yields the following result:
$3w^{2}\frac{\partial x}{\partial w} + 3z^{2}\frac{\partial x}{\partial z} = 5x\frac{\partial y}{\partial x} + y$
After rearranging the second equation, I obtained the following:
$\frac{\partial x}{\partial w} =\frac{( 5x\frac{\partial y}{\partial x}+y - +3z^{2}\frac{\partial x}{\partial z})}{3w^{2}}  $
and taking the partial of the second I get:
$ 2x \frac{\partial x}{\partial w} + 2y \frac{\partial y}{\partial w} + 2z \frac{\partial z}{\partial w} $
 A: The partial derivative $\frac{\partial }{\partial w}$ means deriving with respect to w, that is $\frac{\partial x}{\partial w}$ is the derivative of $x$ as a function of $w$  with respect to $w$ ($x$ is a multivariate function). 
Deriving both sides of $(2)$ with respect to $w$ we get:
$$\frac{\partial (w^3 +z^3) }{\partial w}=\frac{\partial (5xy +12)}{\partial w}$$
By the chain and product rules:
$$3w^2 +3x^2 \frac{\partial z}{\partial w}=5\frac{\partial x}{\partial w}y + 5x\frac{\partial y}{\partial w}$$
Hence, solving for $\frac{\partial x}{\partial w}$:
$$\frac{\partial x}{\partial w}=\frac{3w^2 +3x^2\frac{\partial z}{\partial w}-5x\frac{\partial y}{\partial w}}{5y}$$
A: Note that partial derivatives in a $d$-dimensional environment are only defined when $d$ independent variables have been selected.
Two equations in four variables define a two-dimensional surface $S\subset{\mathbb R}^4$. In the problem at hand the variables $z$ and $w$ are selected as independent variables, and the two equations define $x$ and $y$ implicitly as functions of these, in the following way:
$$x^2(z,w)+y^2(z,w)=6-z^2, \quad 5x(z,w)y(z,w)=z^3+w^3-12\ .$$
Taking ${\partial\over\partial w}$ in both equations we obtain
$$2x(z,w)x_w+2y(z,w)y_w=0,\qquad 5y(z,w)x_w+5x(z,w)y_w=3w^2\ .$$
We now plug in the point $(1,-2,1,1)\in S$ and obtain
$$2 x_w-4y_w=0,\qquad -10 x_w+5y_w=3\ ,$$
from which we easily deduce
$${\partial x\over\partial w}(1,1)=-{2\over5}\ .$$
