To determine the correct statement about the function $f(x) = -2x^8$, we can analyze its first derivative to find critical points and intervals of increase/decrease.

1. **Calculate the first derivative:**
$$f'(x) = \frac{d}{dx}(-2x^8) = -16x^7$$

2. **Find critical points:**
Set $f'(x) = 0$:
$$-16x^7 = 0 \implies x^7 = 0 \implies x = 0$$
The only critical point is $x=0$.

3. **Analyze the sign of $f'(x)$:**
* For $x < 0$, $x^7 < 0$, so $f'(x) = -16(\text{negative number}) > 0$. This means $f(x)$ is increasing for $x < 0$.
* For $x > 0$, $x^7 > 0$, so $f'(x) = -16(\text{positive number}) < 0$. This means $f(x)$ is decreasing for $x > 0$.

4. **Identify the extremum:**
Since $f'(x)$ changes from positive to negative at $x=0$, there is a local maximum at $x=0$. The value of the function at this point is $f(0) = -2(0)^8 = 0$.
Furthermore, for any $x \neq 0$, $x^8 > 0$, so $-2x^8 < 0$. Since $f(0) = 0$ and $f(x) < 0$ for all other $x$, $(0,0)$ is a global maximum.

The correct statement is:
The function $f(x)$ has a global maximum at $x=0$.