Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?

Let $(t,t^4-4t^3)$ be a point of contact, then
$$\frac{dy}{dx} = \frac{\dot{x}}{\dot{y}} = 4t^3-12t^2$$
The equation of tangent is
$$(y-t^4+4t^3)=4t^2(t-3)(x-t)$$
Let $(s^4-4s^3,s)$ be the self-intersection of its tangent.
\begin{align*}
  (s^4-4s^3)-(t^4-4t^3) &= 4t^2(t-3)(s-t) \\
  (s^4-t^4)-4(s^3-t^3) &= 4t^2(t-3)(s-t) \\
  (s^2-t^2)(s^2+t^2)-4(s-t)(s^2+st+t^2) &= 4t^2(t-3)(s-t) \\
  (s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)] &= 4t^2(t-3)(s-t) \\
  (s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)-4t^2(t-3)] &= 0 \\
  (s-t)[s^3+(t-4)s^2+t(t-4)s+t^2(8-3t)] &= 0 \\
  (s-t)^2[s^2+2(t-2)s+t(3t-8)] &= 0 \\
\end{align*}
The double root $s=t$ is due to tangency, thus another intersections are given by $$s=2-t\pm \sqrt{2(2+2t-t^2)}$$
provided $1-\sqrt{3} \le t \le 1+\sqrt{3}$.

The line will be a common tangent if $\Delta=0$,  that is
  $$(s,t)=(1\pm \sqrt{3},1\mp \sqrt{3})$$

