#### Answer

The graph of $f$ has horizontal tangents at $(1,-2)$ and $(-1,2)$.

#### Work Step by Step

$$f(x)=x^3-3x$$
Having a horizontal tangent means that having the slope, and the derivative, there equal $0$.
In other words, we need to find $x$ such that $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=0\hspace{1cm}(1)$$
First, we examine $f(x+h)$:
$f(x+h)=(x+h)^3-3(x+h)=x^3+3x^2h+3xh^2+h^3-3x-3h=(x^3-3x)+(3x^2h+3xh^2+h^3-3h)$
Therefore,
$$\frac{f(x+h)-f(x)}{h}=\frac{3x^2h+3xh^2+h^3-3h}{h}=3x^2+3xh+h^2-3$$
Replacing the result into $(1)$: $$\lim_{h\to0}(3x^2+3xh+h^2-3)=0$$
$$3x^2+3x\times0+0^2-3=0$$
$$3x^2-3=0$$
$$x^2-1=0$$
$$x=\pm1$$
We have $f(1)=1^3-3\times1=1-3=-2$ and $f(-1)=(-1)^3-3\times(-1)=-1+3=2$
Therefore, the graph of $f$ has horizontal tangents at $(1,-2)$ and $(-1,2)$.