This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function

$\displaystyle

f(x) = x^3

$

$\displaystyle f'(x) = 3x^2 $

Line

$\displaystyle

3x-y+1 = 0

$

$\displaystyle y = 3x+1$

From here we know that $\displaystyle m=3$ since $\displaystyle y=mx+b$

Then I made $\displaystyle f'(x)=3$

$\displaystyle 3x^2=3$

$\displaystyle x^2=1$

To get the value of x.

$\displaystyle x=1$

Now I go back to the given line and plug in:

$\displaystyle y=3(1)+1$

$\displaystyle y=4$

My point is $\displaystyle (1,4)$

I did most of it as I was writing this, I think my light bulb turned on.

$\displaystyle y-y1=m(x-x1)$

$\displaystyle y-4=3(x-1)$

$\displaystyle y=3x+1$

Did I do it right or is there another way of doing this?