If $z= \arctan \frac{y}{x}$ show that the following is true $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0$ 
If $z= \arctan \frac{y}{x}$ show that the following is true
  $$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0$$

So I don't truly understand how implicit partial differentiation works, but I understand normal implicit differentiation. I would be great if someone would be able to give some help with this.
The question also gives a side note of 
$$\frac{\partial}{\partial x}\left( \arctan z \right) = \frac{1}{1+z^2}\times \frac{\partial z}{\partial x}$$
Im thinking you start by putting the euqation as $\tan z =\frac{y}{x}$?
 A: Let $u=y/x$ so $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=x\cdot\frac{-y}{x^2}+y\cdot\frac{1}{x}=0$. Then $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{dz}{du}\cdot 0=0$.
A: $$\frac{\partial z}{\partial x}=\frac{y}{1+\frac{y^2}{x^2}}\cdot (\frac{-1}{x^2})=\frac{-y}{x^2+y^2}$$ 
and 
$$\frac{\partial z}{\partial y}=\frac{1}{1+\frac{y^2}{x^2}}\cdot \frac1x=\frac{x}{x^2+y^2}.$$
Now substitute in
$$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}.$$
A: Just for the fun of it, and for the sake of "completeness", let's first find the derivative of $\arctan w$ ourselves:
Set
$z = \arctan w; \tag 1$
then
$w(z) = \tan z = \dfrac{\sin z}{\cos z}; \tag 2$
we have
$w'(z) = \dfrac{(\cos z)(\cos z) - (-\sin z)(\sin z)}{\cos^2 z} = \dfrac{\cos^2 z + \sin^2 z}{\cos^2 z} = \dfrac{1}{\cos^2 z} = \sec^2 z; \tag 3$
we use the identity
$1 + \tan^2 z = 1 + \dfrac{\sin^2 z}{\cos^2 z} = \dfrac{\cos^2 z + \sin^2 z}{\cos^2 z} = \dfrac{1}{\cos^2 z} = \sec^2 z \tag 4$
to write
$\dfrac{dw}{dz} = w'(z) = 1 + \tan^2 z = 1 + w^2, \tag 5$
whence
$z'(w) = \dfrac{dz}{dw} = \dfrac{d(\arctan w)}{dw} = \dfrac{1}{1 + w^2}. \tag 6$
Now having $z'(w)$ at hand, we set
$w = \dfrac{y}{x} = yx^{-1}, \tag 7$
and use the chain rule to find $\partial z / \partial x$, $\partial z / \partial y$:
$\dfrac{\partial z}{\partial x} = \dfrac{dz}{dw} \dfrac{\partial w}{\partial x}, \tag 8$
$\dfrac{\partial w}{\partial x} = \dfrac{\partial (yx^{-1})}{\partial x} = -yx^{-2}; \tag 9$ 
we fold (6) and (9) into (8):
$\dfrac{\partial z}{\partial x} = -yx^{-2}\dfrac{1}{1 + w^2} , \tag{10}$
also,
$\dfrac{\partial w}{\partial y} = \dfrac{\partial (yx^{-1})}{\partial y} = x^{-1}, \tag{10}$ 
whence,
$\dfrac{\partial z}{\partial y} = \dfrac{dz}{dw} \dfrac{\partial w}{\partial y} = x^{-1}\dfrac{1}{1 + w^2}; \tag{11}$
therefore,
$x \dfrac{\partial z}{\partial x} + y \dfrac{\partial z}{\partial y} = -yx^{-1}\dfrac{1}{1 + w^2} + yx^{-1}\dfrac{1}{1 + w^2} = 0, \tag{12}$
as was to be proved.
So much for the "standard derivation" based upon straightforward partial differentiation.  However,
There is in fact a much swifter, easier way to do this: 
We have the radial vector
$\mathbf r = \begin{pmatrix} x \\ y \end{pmatrix}, \tag{13}$
and that the central angle which $\mathbf r$ makes with the $x$-axis is
$\theta = \arctan \dfrac{y}{x} = z; \tag{14}$
it follows that
$\nabla \theta = \begin{pmatrix} \dfrac{\partial \theta}{\partial x} \\ \dfrac{\partial \theta}{\partial y} \end{pmatrix} = \begin{pmatrix} \dfrac{\partial z}{\partial x} \\ \dfrac{\partial z}{\partial y} \end{pmatrix}; \tag{15}$
thus,
$\mathbf r \cdot \nabla \theta = \nabla_{\mathbf r} \theta = 0, \tag{16}$
since $\theta$ does not change in the $\mathbf r$ direction; indeed, we see that, for $0 < \alpha \in \Bbb R$,
$\arctan \dfrac{\alpha y}{\alpha x} = \arctan \dfrac{y}{x}, \tag{17}$
which shows that $\theta$ is invariant along the rays $\alpha \mathbf r$.  Therefore,
$x \dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = \mathbf r \cdot \nabla \theta = \nabla_{\mathbf r} \theta = 0, \tag{18}$
as required.  $OE\Delta$.
A: Here is a way to do this without implicit differentiation.
$$x\frac{\partial z}{\partial x}=\frac{x}{1+(yx^{-1})^2}\frac{-1}{x^2}=-\frac{1}{x(1+(yx^{-1})^2)}$$
$$y\frac{\partial z}{\partial y}=\frac{x^{-1}}{1+(yx^{-1})^2}=\frac{1}{x(1+(yx^{-1})^2)}$$
Therefore,
$$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$$
A: first apply the chain rule to evaluate the partial derivative.
$$\frac{\partial z}{\partial x}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x^2}\right)$$
$$\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(\frac{1}{x}\right)$$
so,now
$$x\frac{\partial z}{\partial x}=\frac{x}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x^2}\right)=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x}\right)......(1)$$
$$y\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(\frac{y}{x}\right)..................(2)$$
adding (1)+(2),
$$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left[\left(-\frac{y}{x}\right)+\left(\frac{y}{x}\right)\right]=0$$
