Implicit differentiation of an inverse-tangent/logarithmic equation

Differentiate the left side: with $u = y/x$, $\frac{d}{dx}\arctan(u) = \frac{1}{1+u^2}\cdot u'$, and by the quotient rule $u' = \frac{y'x - y}{x^2}$. Since $1+u^2 = 1+\frac{y^2}{x^2} = \frac{x^2+y^2}{x^2}$, the left side becomes $\frac{x^2}{x^2+y^2}\cdot\frac{xy'-y}{x^2} = \frac{xy'-y}{x^2+y^2}$.

Differentiate the right side using the chain rule: $\frac{d}{dx}\ln(x^2+y^2) = \frac{2x+2y y'}{x^2+y^2}$. Setting the two derivatives equal and noting both sides share the denominator $x^2+y^2$, we get $xy' - y = 2x + 2yy'$. Solving for $y'$: $xy' - 2yy' = 2x + y$, so $y'(x-2y) = 2x+y$, giving $y' = \dfrac{2x+y}{x-2y}$.

The shared denominator was x squared plus y squared. This common factor is exactly why it could be cancelled from both sides, leaving the simpler equation xy prime minus y equals two x plus two y y prime, which was then solved for dy/dx.