I got the second derivative $12x-6$ for this equation $y = 2 x ^ { 3 } - 3 x ^ { 2 } - 36 x + 4 = 0$ and the local minimum is something $3$ and $-2$ as local maximum, but I don't know how to find absolute maximum and minimum.
Am I supposed to use those interval points as $f(0)$ and $f(4)$ and apply to derivative and answer should be $-6$ as absolute maximum and $42$ and absolute minimum. I don't know. Please guide me.
Guide:
Your derivative is not correct. You have computed the second derivative instead.
$$f'(x) = 6x^2-6x-36=0$$
To find relative maxima and minima, set $f'(x) =0$ and solve for $x$. Those are candidate for relative maxima and minima. You can verify which one is it by evaluating the second deritive.
To find absolute maxima and minima in $[0,4]$, compare the values of $f(0), f(4)$, as well as any values for relative maxima and minima that you have obtained earlier if they reside in the interior.