First Order ODE/x dy = (y + x^2 + 9 y^2) dx/Proof 2

Proof
Let $z = \arctan \left({3y / x}\right)$.

Then:
 * $\dfrac {\partial z} {\partial x} = \dfrac 1 {1 + \left({3 y / x}\right)^2} \dfrac {-3 y} {x^2} = \dfrac {-3 y} {x^2 + 9 y^2}$


 * $\dfrac {\partial z} {\partial y} = \dfrac 1 {1 + \left({3 y / x}\right)^2} \dfrac 3 x = \dfrac {3 } {x^2 + 9 y^2}$

So:
 * $\mathrm d z = \dfrac {3 x \, \mathrm d y - 3 y \, \mathrm d x} {x^2 + 9 y^2}$

Multiplying $(1)$ by $3$ and manipulating:
 * $\dfrac {3 x \, \mathrm d y - 3 y \, \mathrm d x} {x^2 + 9 y^2} = 3 \, \mathrm d x$

From Differential of Arctangent of Quotient:
 * $\mathrm d \left({\arctan \left({\dfrac {3 y} x}\right)}\right) = 3 \, \mathrm d x$

and so
 * $\arctan \left({\dfrac {3 y} x}\right) = 3 x + C$