Showing a result using partial derivatives, given that $f(u,v)=0$ The question I am attempting states:
Show that if $f(u,v)=0$, where $u=x+y$ and $v=x^2+xy+z^2$, then:
$$x+y=2z\left(\left(\frac{\partial z}{\partial y}\right)_x-\left(\frac{\partial z}{\partial x}\right)_y\right) \qquad (1)$$
My working so far is:
$$ \left(\frac{\partial v}{\partial x}\right)_y=2x+y+2z\left(\frac{\partial z}{\partial x}\right)_y $$
and
$$\left(\frac{\partial v}{\partial y}\right)_x = x+2z\left(\frac{\partial z}{\partial y}\right)_x $$
and so:
$$\left(\frac{\partial v}{\partial y}\right)_x-\left(\frac{\partial v}{\partial x}\right)_y=-x-y+2z\left(\left(\frac{\partial z}{\partial y}\right)_x-\left(\frac{\partial z}{\partial x}\right)_y\right) \qquad (2)$$
I can see that this leads to (1) if the LHS is zero, but I'm not sure how to show it. I assume it must involve $f(u,v)=0$ and $u=x+y$, since I haven't used those bits of information yet, but again, I'm not sure how to apply them. I tried using the cyclic relation to give:
$$ \left(\frac{\partial v}{\partial y}\right)_x-\left(\frac{\partial v}{\partial x}\right)_y\equiv \left(\frac{\partial v}{\partial y}\right)_x\left(1+\left(\frac{\partial y}{\partial x}\right)_v\right)$$
which could give zero if $ \left(\frac{\partial v}{\partial y}\right)_x=0 $, but I can't see how that could be, given its presence in the result to be proved. Equally I can't see how $ \left(\frac{\partial y}{\partial x}\right)_v=-1 $. 
Alternatively, I tried:
$$ \left(\frac{\partial f}{\partial x}\right)_y = \left(\frac{\partial f}{\partial u}\right)_v \left(\frac{\partial u}{\partial x}\right)_y + \left(\frac{\partial f}{\partial v}\right)_u \left(\frac{\partial v}{\partial x}\right)_y \qquad (3)$$which could then be rearranged to give:
$$ \left(\frac{\partial v}{\partial x}\right)_y = \left(\frac{\partial v}{\partial f}\right)_u \left(\left(\frac{\partial f}{\partial x}\right)_y - \left(\frac{\partial f}{\partial u}\right)_v \right)$$
since $\left(\frac{\partial u}{\partial x}\right)_y=1$. Combined with the corresponding result for $\left(\frac{\partial v}{\partial y}\right)_x$ and substituted into the LHS of (2) this gives:
$$\left(\frac{\partial v}{\partial y}\right)_x-\left(\frac{\partial v}{\partial x}\right)_y=\left(\frac{\partial v}{\partial f}\right)_u \left( \left(\frac{\partial f}{\partial y}\right)_x-\left(\frac{\partial f}{\partial x}\right)_y \right) $$
And this leaves me at a block again.
Could anyone give me a hint? As well, is the expansion in (3) even valid – i.e can I hold just y constant on the LHS, or do I have to do it with both y and z constant? I'm rather new to multivariable stuff and I feel like I'm not understanding something fundamental? 
 A: I think you have the idea, but I usually draw a tree diagram to visualize the dependence between the variables first when I studied multi var last year. It looks to me that it shall be like this (just one way to draw such a diagram, some other textbooks might draw that differently):

Since $f(u(x,y)),v(x,y,z(x,y))) = 0$, $\frac{\partial f}{\partial x}= \frac{\partial f}{\partial y}= 0$, since they are the independent variables [see the remarks for elaboration]. All that remains is the "computational work". To find $\frac{\partial}{\partial x}$, we have to trace the branches that have $x$ at the end.

\begin{align}\frac{\partial f}{\partial x}&=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial z}\frac{\partial z}{\partial x}\\
0 &= \frac{\partial f}{\partial u} 1 + \frac{\partial f}{\partial v} (2x + y) + \frac{\partial f}{\partial v} (2z) \frac{\partial z}{\partial x} \tag{1}\end{align}
Similarly for $\frac{\partial}{\partial y}$,
\begin{align}\frac{\partial f}{\partial y}&=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial z}\frac{\partial z}{\partial y}\\
0 &= \frac{\partial f}{\partial u} 1 + \frac{\partial f}{\partial v} x + \frac{\partial f}{\partial v} (2z) \frac{\partial z}{\partial y} \tag{2}\end{align}
If we subtract $(1)$ by $(2)$, we obtain
$$0 = \frac{\partial f}{\partial v}(2x+y-x) + \frac{\partial f}{\partial v}(2z) \bigg(\frac{\partial z}{\partial x} -\frac{\partial z}{\partial y}  \bigg)$$
Assuming $\frac{\partial f}{\partial v} \neq 0$, we cancel that out. Rearranging we get
$$x+y = (2z) \bigg(\frac{\partial z}{\partial y}-\frac{\partial z}{\partial x}  \bigg)$$

In the answer, I wrote $f(u(x,y), v(x,y,z(x,y)))$ instead of $f(u,v)$ was to show that in this case, $f$ is really a function of $(x,y)$. In fact, it is clearer for us to write $g(x,y) =f(u(x,y), v(x,y,z(x,y)))= 0$ for all $(x,y)$. Then by the definition of partial derivative, 
$$\frac{\partial g}{\partial x} =  \lim_{h \to 0} \frac{g(x+h,y)- g(x,y)}{h} = 0$$ 
, since the numerator always vanishes.
