That was indeed a tricky question!
First of all, you need to differentiate the equation so you can find the equation of the gradient (of the tangent lines) for this graph:
Secondly, you start deliberating: as our tangent lines pass through the origin (0,0), you can conclude that the equations of these lines should be of the form y=kx where k is a gradient of these tangent lines. Therefore k=y/x. Indeed, the gradient of any line which goes through the origin can be calculated as the y-coordinate divided by the x-coordinate of any point of the line.
Now, we know that the lines are tangent lines to our equation y=x^3 - 3x^2 - 4x. They touch the graph of this function at the points A and B. So the coordinates of these points can be written as
(x, y=x^3 - 3x^2 - 4x). Therefore:
k = y/x = (x^3 - 3x^2 - 4x)/x = x^2 - 3x - 4
From differentiation you also know that the gradient (of the tangents):
k = f’(x) = 3x^2-6x-4
So you can conclude that x^2 - 3x - 4 = 3x^2 - 6x - 4
Solving this quadratic equation you find that:
x = 0
x = 1.5
These are the x-coordinates of the points A and B. To find the y-coordinates of these points you have to substitute these values into the original equation, so you will have two points: (0,0) and (1.5, -9.375).
Lastly, you need to find the equations of these tangent lines. You already know that they can be described using the equation y=kx where k is the gradient. To find the gradient of the tangent lines at the points A and B you substitute the x-coordinates of these points into the differentiated equation:
f’(0)= -4 so the tangent line is y = -4x
f’(1.5) = -6.25 so the tangent line is y = -6.25x
Below is the graph of this function with both tangents:
I focused on the main steps, skipping some calculations. If you have any questions feel free to ask.