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 $x_0 = \pm 1$.
You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point. The vertical line does that and you've already found it. Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).
Nonvertical lines passing through the point $(4,6)$ have equation $y-6=m(x-4)$.
That implies $y=mx-4m+6$, so we can put $mx-4m+6$ in place of $y$:
$$
\frac{x^2}{16} + \frac{(mx-4m+6)^2}{4} = 1.
$$
This is equivalent to
$$
\underbrace{(1+4m)}x^2 + \underbrace{-8m(4m-6)}\ x + \underbrace{4(4m-6)^2 -16} = 0.
$$
This equation is quadratic in $x$. We therefore want a quadratic equation with exactly one solution. A quadratic equation $ax^2+bx+c=0$ has exactly one solution precisely if its discriminant $b^2-4ac$ is $0$. So we have
$$
b^2-4ac = \underbrace{64m^2(4m-6)^2 - 4(1+4m)(4(4m-6)^2-16) = 0}.
$$
Now we only need to solve this last equation for $m$.
Best Answer
1) Find slopes of the tangents to the ellipse at any point on the ellipse. Implicit differentiation helps here, i.e.
$$2 x + 8 y \frac{dy}{dx} = 0$$
2) Find the point(s) on the ellipse whose tangent passes through $(4,6)$. The equation of a line is
$$y-y_0 = m (x - x_0)$$
where $m$ is the slope at the point $(x,y)$ on the ellipse and $(x_0,y_0)$ is the point $(4,6)$. You end up with an equation like
$$y-6 = \frac{dy}{dx} (x-4)$$
Plug in the derivative for the slope; you will get 2 equations and 2 unknowns (the other equation being the equation for the ellipse). Solution will get you the point(s), and thus the slope(s) and the equation(s) of the lines.
So you can check your results, I get two points: $(4,0)$ (obvious in hindsight) and $(-16/5,6/5)$.