When thinking about stationary points, it is important to remember that these points happen when the gradient of the curve is 0 (-> dy/dx = 0), or if you prefer, when the rate of change of a graph is 0. 

So the first step is to differentiate the given equation:

If we take the first part of the equation, 6ln(x), we can split this into two seperate parts, ln(x) and 6, where these variables are multiplied together. Using the chain rule, where u = 6 and v = ln(x), we can calculate the differential by calculating v.du + u.dv, where dv is the differential of v, and du is the differential of u.

As the differential of ln(x) is 1/x, and the differential of 6 = 0, v.du is equal to ln(x).0, and u.dv is equal to 6.(1/x) which is equal to 6/x.

Next we move on to the differentiation of x² , and as differentating involves multiplying by the current power, and then taking away 1 from the previous power, we get 2x. 

Next, 8x differentiates to 8, as x¹ multiplied by 1 and minus 1 is equal to x0 , and 8.1 = 8.

Finally, 3 differentiates to 0, as the derivative represents the rate of change, and a constant factor has a rate of change of 0 (as it is constant).

This leaves us with: 

dy/dx = 6/x + 2x - 8

Getting rid of the fraction by multiplying by x, we get:

dy/dx = 6 + 2x² - 8x (which rearranges to give)

-> dy/dx = 2x² -8x + 6 

By factorising the equation out, we get that x is either equal to 1 or 3:

= 2x² - 8x + 6

= (2x - 6)(x-1)

-> 2x - 6 = 0 or x - 1 = 0

-> x = 3 or x = 1

Substitute in the values we have calculated into the original equation:

x = 1

y = 6ln(1) +1² - (8.1) + 3        (ln(1) = 0)

y = 1 - 8 + 3

y = -4

So when x = 1, y = -4

x = 3

y = 6ln(3) + 3² - (8.3) + 3

y = 6ln(3) + 9 - 24 + 3

y = 6ln(3) - 12 

So when x = 3, y = 6ln(3) - 12

So the coordinates of the stationary points are:

(1, -4) and (3, 6ln(3) - 12)