Find the slope intercept equation of the line normal to the graph of y=x^3 at the point at which x=1/3?

Here is what I have:

First, (f(x+h)-f(x))/h= ((x+h)^3-x^3)/h= 3x^2+3hx+h

lim(h-->0) 3x^2+3xh+h^2=3x^2

y=3x^2

Then, I plugged 1/3 into the y=3x^2

y=3*(1/3)^2
y=1/3

Then I start plugging in point

(1/3,f(1/3))=(1/3,1/27)

to complete the y=mx+b...

y=1/3x+b

find b by plugging in (1/3,1/27)
1/27=1/3*1/3+b
1/27=1/9+b
1/27-1/9=b
-2/27=b
y=1/3-2/27

is that alright?