Find the points on the curve y=x^3 at which the normal has a gradient of -3/4?

"normal" in this context means "perpendicular to". If the normal q has a gradient (slope) m of -3/4, then the slope m_t of the tangent t is the negative reciprocal of m_q:

m_t = - (m_q)^-1 = 4/3.

The slope of the tangent t of a function f (x) at x_0 is given by the 1st derivative of f (x):

f (x) = x^3

f '(x) = 3x^2.

Thus, m_t = f '(x_0) = 3(x_0)^2.

We know that m_t = 4/3. So

3x^2 = 4/3
x^2 = 4/9

x_1,2 = +- 2/3.

Find f(x_1) and f(x_2), and you're done:

f(2/3) = (2/3)^3 = 8/27 , and

f( -2/3) = ( -2/3)^3 = - 8/27.

Thus, the points on the curve y=x^3 at which the normal has a gradient of -3/4 are:

P_1 (2/3 , 8/27), P_2 ( -2/3, -8/27).