Answer
If $f$ is differentiable at $x=x_0$ then $-f$ is also differentiable at $x=x_0$.
Work Step by Step
$f$ is differentiable at $x=x_0$ means that $$\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$$ has a defined value.
Now about $-f$, the derivative of $-f$, $(-f)'$, at $x=x_0$ has the following formula:
$$(-f)'=\lim_{h\to0}\frac{(-f)(x_0+h)-(-f)(x_0)}{h}=\lim_{h\to0}\frac{-f(x_0+h)-(-f(x_0))}{h}$$
$$(-f)'=\lim_{h\to0}-\frac{f(x_0+h)-f(x_0)}{h}=-\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$$
Since $\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$, as said above, has a defined value, $(-f)'=-\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$ also has a defined value.
This means $-f$ is also differentiable at $x=x_0$.
