How do I find f ' (3)

How would I find f ' (3) using IMPLICIT DIFFERENTIATION?

Let y = f(x). If xy^3 + xy = 6, and f(3) = 1, find f ' (3)

Can someone please explain or show steps so I can understand? Thank you

Differentiate both sides implicitly

d(xy^3)+d(xy)=0

Use the chain rule

y^3+3xy^2y'+y+xy'=0

Plug in y=f(3)=1, and find y'=f'(3)....

1+9y'+1+3y'=0
y'=f'(3)=-1/6

f(x)=y
So f'(x)=y,actually x and y is y and x..

So f'(3)=?,y=3
x.3^3+x.3=6
27x+3x=6
30x=6
x=1/5