Find the equation of the tangent line to the graph of (x+y)^3=27x at the point (1,2)

Answer 1

take the differential
3(x+y)^2 (1+y')=27
y'+1=9/(x+y)^2

y'=9/(x+y)^2 -1 = m

this is the slope

y=mx+b at the point 1,2
solve for b
when x=1, y=2
then the equation is
y=mx + b

Answer 2

I understand , but when you plug in the x and y in and I get y' =0 so would that be my slope?