Differentiation

(x/3 ) * (dy/dx) - (1/9)y^2 = 1 (x>0)
(x/3 ) * (dy/dx) = 1+(1/9)y^2
9x (dy/dx) = 27+3y^2
3x (dy/dx) = 9+y^2
dy/(9+y^2)=dx/3x
∫1/(9+y^2) dy=∫1/3x dx
(1/3)arctan(y/3)=(1/3)lnx+C
Sub x=1, y=π
(1/3)√3=C
C=1/√3
So the particular solution is
(1/3)arctan(y/3)=(1/3)lnx+1/√3
arctan(y/3)=lnx+√3
Note:
For calculate ∫1/(9+y^2) dy
Let y=3tanθ ; dy=3sec^2θ dθ
∫1/(9+y^2) dy
=∫[1/(9+9tan^2θ)] (3sec^2θ)dθ
=∫sec^2θ/3(sec^2θ) dθ
=(1/3)θ+C
=(1/3)arctan(y/3)+C