The derivative of the function is $4x^3 - 12x$, so the slope of the tangent line through the point $(x_0, f(x_0))$ will be exactly $4x_0^3 - 12x_0$. We may actually write the line in slope-intercept form as
$$y - f(x_0) = (4x_0^3 - 12x_0)(x - x_0)$$
or alternatively,
$$y = (4x_0^3 - 12x_0)(x - x_0) + x_0^4 - 6x_0^2$$
So the question is asking under what conditions on $x_0$ the point $(0, 3)$ lies on this curve. Substituting the values, we find that
$$3 = (4x_0^3 - 12x_0)(0 - x_0) + x_0^4 - 6x_0^2$$
Rearranging, this leads to
$$-3x_0^4 +6x_0^2 - 3 = 0$$
Dividing by $-3$ leads to
$$x_0^4 - 2x_0^2 + 1 = 0 \implies (x_0^2 - 1)^2 = 0$$
Hence, we see that $x_0^2 = 1$, so that $x_0 = \pm 1$.
Best Answer
Expand the factors of $f(x)$. You can use the product rule to differentiate as is, but it would be simpler to just expand the factors, then differentiate. $$f(x) = (2x - 1)(x+1) = 2x^2 +x - 1$$
Find the intercepts the original curve $f(x)$: Solve for the x-values where $f(x) = 0$: there will be two solutions in $x$: $x_1, x_2$. $$\begin{align}f(x) = 0 & \iff (2x - 1)(x + 1) = 0 \\ \\ & \iff 2x - 1 = 0 \;\text{ or } \; x + 1 = 0 \\ \\ &\iff x_1 = \dfrac 12, \;\text{ or }\; x_2 = -1\end{align}$$
Find $f'(x)$.