Locating the critical point

Differentiate: $f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2$. Setting $f'(x) = 0$ gives $3(x-1)^2 = 0$, so $x = 1$ is the only solution.

Since $(x-1)^2 \ge 0$ always, $x=1$ is the unique critical number, and it is a repeated (double) root of $f'$.

From $f'(x) = 3x^2 - 6x + 3$, differentiate again: $f''(x) = 6x - 6$.

At $x = 1$: $f''(1) = 6(1) - 6 = 0$. Since $f''(1) = 0$, the Second Derivative Test is inconclusive at this critical point — it gives no information about whether $x=1$ is a local max, local min, or neither.

Since $f'(x) = 3(x-1)^2 \ge 0$ for all $x$, the derivative does not change sign at $x=1$: it is positive on both sides of $x=1$ (and zero only at $x=1$ itself).

Because $f'$ does not switch from positive to negative or negative to positive, $x=1$ is neither a local maximum nor a local minimum. Indeed $f(x) = (x-1)^3$, which is monotonically increasing with an inflection point (not an extremum) at $x=1$.