Answer
$-f'(x_0)$ also exists.
Work Step by Step
Step 1. Assume we know $f(x)$ is differentiable at $x=x_0$, which means that $f'(x_0)=\lim_{x\to x_0}\frac{f(x+x_0)-f(x)}{x-x_0}$ exist.
Step 2. For a new function defined as $g(x)=-f(x)$, try to find the derivative of $g(x)$ at $x=x_0$ as:
$g'(x_0)=\lim_{x\to x_0}\frac{g(x+x_0)-g(x)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x+x_0)+f(x)}{x-x_0}=-f'(x_0)$
also exists.
