I'd love to get some help understanding this question.
Let $ w=f(x,y,z) $ with the constraint $g(x,y,z)=3.$ At point $P(0,0,0)$ we have $df = <1,1,2> , dg=<2,-1,-1>.$ Find the value of P at two quantities:
$$\left(\frac{\partial z}{\partial x}\right)_y$$ and $$\left(\frac{\partial w}{\partial x}\right)_x$$
[Attempt]
Rewriting G as $$G(x,y)=g[x,y;z(x;y)]=3$$
and applying the chain rule:
$$\frac{\partial z}{\partial x} = - \frac{\frac{\partial G}{\partial x}}{\frac{\partial G}{\partial z}}=- \frac{G_x}{G_z} = - \frac{2}{-1} = + 2 $$
Given the m.v.c. rule: $$ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial u}$$
Rewriting w and re-applying the chain rule using the previous result: $$w = f(x,y,z(x,y))$$ $$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \frac{\partial z}{\partial x} + 0 $$
$$ \left(\frac{\partial w}{\partial x}\right)_y = \frac{ \partial w}{\partial z}{\frac{\partial z}{\partial x}} = 2 \hspace{0.1cm} 2 = 4$$
HINT: By the implicit function theorem, the equation $g(x,y,z)=3$ defines $z=z(x,y)$ near $P$. Use the chain rule to differentiate $G(x,y) = g(x,y,z(x,y))$ with respect to $x$. Note that $G(x,y)=3$ for all $x,y$ near $(0,0)$.
Similarly, for the second problem [once it's modified], you want to differentiate $F(x,y) = f(x,y,z(x,y))$ [for the same $z(x,y)$] using the chain rule. In particular, you will have $$\left(\frac{\partial w}{\partial x}\right)_y = \frac{\partial F}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}.$$