Find $y'$ and $y''$ $\log(\sqrt{x^2 + y^2})=\arctan{\frac{y}{x}}$
Let $F(x,y)= \log(\sqrt{x^2 + y^2})-\arctan{\frac{y}{x}}$
I can't do it right, because I don't understand the formula $\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$
I think $z=y(x)$. So should I get something like that?
$$\frac{\partial F}{\partial x}=\frac{x}{x^2+y^2}$$  and
$$\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=\frac{y\cdot y'}{x^2+y^2}$$    

Frankly, I would just differentiate directly, using the chain rule, on both sides.  And, before differentiating, reduce the left side: $\log(\sqrt{x^2+ y^2})= \frac{1}{2}\log(x^2+ y^2)$
The derivative of $\frac{1}{2}\log(x^2+ y^2)$ is 
$$
\frac{1}{x^2+ y^2}(x+ yy')= \frac{x+ yy'}{x^2+ y^2}.
$$
The derivative of $\arctan(\frac{y}{x})$ is 
$$
\frac{1}{1+\left(\frac{y}{x}\right)^2}
\left(-\frac{y}{x^2}+ \frac{1}{x}y'\right)= \frac{xy'- y}{y^2y'}.
$$
Now solve for $y'$ in
$$
\frac{x+ yy'}{x^2+ y^2}= \frac{xy'- y}{y^2y'}.
$$
