<s>[INST] 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:
$\dfrac{1}{(1+(y/x)^2)}*(\dfrac{y}{x})'=(x^2+y^2)^\dfrac{-1}{2}*(\dfrac{1}{2})*(x^2+y^22)^\dfrac{-1}{2}  * (2x+2y'*y')$
 [/INST] $$\arctan(\frac{y}{x})=\ln\sqrt{x^2+y^2}$$
$$\frac{1}{1+(y/x)^2}\frac{y'x-y}{x^2}=\frac{x+yy'}{\sqrt{x^2+y^2}}$$
solve for $y'$
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