$dy/dx = (9x + 4y + 1)^2$
$dy = (9x + 4y + 1)^2~dx$
Let
$z = 9x + 4y + 1$
$dz = 9~dx + 4~dy$
$dy = \frac{1}{4}(dz - 9~dx)$
$\frac{1}{4}(dz - 9~dx) = z^2~dx$
$dz - 9~dx = 4z^2~dx$
$dz = 4z^2~dx + 9~dx$
$dz = (4z^2 + 9)~dx$
$\dfrac{dz}{4z^2 + 9} = dx$
$\dfrac{dz}{(2z)^2 + 3^2} = dx$
$\displaystyle \dfrac{1}{2} \int \dfrac{2dz}{(2z)^2 + 3^2} = \int dx$
$\dfrac{1}{2} \left[ \dfrac{1}{3} \arctan \left( \dfrac{2z}{3} \right) \right] = x + \frac{1}{6}c$
$\frac{1}{6} \arctan (\frac{2}{3}z) = x + \frac{1}{6}c$
$\arctan (\frac{2}{3}z) = 6x + c$
$\frac{2}{3}z = \tan (6x + c)$
$2z = 3\tan (6x + c)$
But $z = 9x + 4y + 1$
$2(9x + 4y + 1) = 3\tan (6x + c)$
$3\tan (6x + c) = 2(9x + 4y + 1)$ answer
