What is the right solution for derivative of implicit function? There's need to find y' of:
$$\arctan(y/x)=\ln\sqrt{x^2 + y^2}$$
Tried:
$\dfrac{1}{(1+(y/x)^2)}*(\dfrac{y}{x})'=(x^2+y^2)^\dfrac{-1}{2}*(\dfrac{1}{2})*(x^2+y^22)^\dfrac{-1}{2}  * (2x+2y'*y')$

$$\arctan \frac yx=\ln\sqrt{x^2+y^2}=\frac12\ln(x^2+y^2)$$
Using Chain Rule, $$\frac{d \left(\arctan \frac yx\right)}{d\left(\frac yx\right) }\cdot \frac{d\left(\frac yx\right) }{dx}=\frac12\frac{d\left(\ln(x^2+y^2)\right)}{d(x^2+y^2)}\cdot\frac{d(x^2+y^2)}{dx}$$
$$\frac1{1+\frac{y^2}{x^2}}\frac{x\frac{dy}{dx}-y}{x^2}=\frac1{2(x^2+y^2)}\left(2x+2y\frac{dy}{dx}\right)$$
Now simplify and take out $\frac{dy}{dx}$
