Two perpendicular lines have slopes that are negative reciprocals of each other. Another way to say that is the product of their slopes is -1. So what is the slope of the first line? Solve for y and the co-efficient of x is the slope.

3x - y = 4

-y = 4 - 3x

y = 3x - 4

The slope of the line is 3. Therefore the slope of the tangent line will be -1/3.

y = x^3 - 3x - 4

To determine the slope of the tangent line, take the derivative. We really just need the power rule.

dy/dx = 3x^2 - 3

So where does the slope of the tangent line equal -1/3?

-1/3 = 3x^2 - 3

Multiply everything by 3 to get rid of the fraction.

-1 = 9x^2 - 9

9x^2 - 9 + 1 = 0

9x^2 - 8 = 0

x^2 = 8/9

x = +- sqrt(8) / 3 = +- 2sqrt(2) / 3 ~= 0.9428

Put those x-values into the equation of the curve to get

y = (+sqrt(8) / 3)^3 - 3(+sqrt(8) / 3) - 4

y = 16sqrt(2) / 27 - 2sqrt(2) - 4

y = (16/27 - 2)sqrt(2) - 4

y = (-38 / 27)sqrt(2) - 4 

y ~= -5.990

y = (-sqrt(8) / 3)^3 - 3(-sqrt(8) / 3) - 4

y = -16sqrt(2) / 27 + 2sqrt(2) - 4

y = (-16/27 + 2)sqrt(2) - 4

y = (38 / 27)sqrt(2) - 4 

y ~= -2.010

So the points on the curve are (+sqrt(8)/3, (-38 / 27)sqrt(2) - 4) and (-sqrt(8)/3, (38 / 27)sqrt(2) - 4) . To find the equation, use the point-slope form of the equation of a line.

y - y1 = m(x - x1), where (x1, y1) is a point on the line.

Just fill in for each point in (x1, y1) and simplify to solve.