Zero of the function!

 'f(x)=1/2*(x^3-3x^2+6)

So we have

(1/2) (x^3 - 3x^2 + 6)  = 0        divide through by 1/2

x^3 - 3x^2 + 6  = 0      the possible rational zeroes are ±1, ± 2, ± 3 and ± 6

Unfortunately........none of these produce a "zero"

If you know Calculus, you could adopt an iterative approach [ Newton's Method] to get an approximate answer or answers

Another alternative is to just graph the function and see where the zeroes lie.......this is guaranteed to have one "real" zero

Here's the graph: https://www.desmos.com/calculator/dx17hd5ze5

WolframAlpha  gives the real solution at  x ≈ -1.1598

Here is the Wolframalpha answer and graph : https://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+%2B+6+%C2%A0%3D+0

There are also two non-real solutions of  x ≈ 2.0979 + 0.7850 i    and  x ≈ 2.0979 - 0.7850 i

If it's just the real zero you are interested in, start by plotting a graph:

Using a graphing calc I get   -1.196  as a zero

Correction to my previous answer....the real root occurs at  x ≈ -1.1958

etc