Answer
There are 2 tangent lines satisfying the requirements: $(l_1):y=3x-3$ and $(l_2):y=3x-7$.
Work Step by Step
$$(s): y=x^3-3x^2+3x-3$$ $$(d): 3x-y=15$$
1) Find the derivative of $y$ of curve $(s)$: $$y'=\frac{d}{dx}(x^3)-3\frac{d}{dx}(x^2)+3\frac{d}{dx}(x)-\frac{d}{dx}(3)$$ $$y'=3x^2-6x+3$$
2) The slope of the tangent line $(l)$ of curve $(s)$ at point $A(a,b)$ equals $y'(a)$.
Consider line $(d)$: $$(d): 3x-y=15$$ $$(d):y=3x-15$$
So the slope of line $(d)$ equals 3.
For 2 lines to be parallel with each other, their slopes must be equal. That means, for the tangent line $(l)$ and line $(d)$ to be parallel with each other, $$y'(a)=3$$ $$3a^2-6a+3=3$$ $$3a^2-6a=0$$ $$3a(a-2)=0$$ $$a=0$$ or $$a=2$$
- For $a=0$, then $b=0^3-3\times0^2+3\times0-3=-3$
- For $a=2$, then $b=2^3-3\times2^2+3\times2-3=-1$
Therefore, there are 2 points $A(0,-3)$ and $B(2,-1)$ at which the tangent line is parallel with line $(d)$.
*Point $A(0,-3)$
The equation of the tangent line $(l_1)$ at $A$ would be $$(l_1):(y+3)=3(x-0)$$ $$(l_1):y=3x-3$$
*Point $B(2,-1)$
The equation of the tangent line $(l_2)$ at $B$ would be $$(l_2):(y+1)=3(x-2)$$ $$(l_2):y+1=3x-6$$ $$(l_2):y=3x-7$$
