The derivative is continuous everywhere; that means that it cannot change signs without going through $0$, by the Intermediate Value Theorem. Since the only places where the derivative is zero are at $x=-1$, $x=0$, and $x=1$, that means that those are the only places where the derivative could, possibly, change signs.
Why do we care about where the derivative could change signs? Because the sign of the derivative tells you whether the original function is increasing or decreasing.
So... what is the sign on the derivative on $(-\infty,-1)$? Plugging in $x=-2$ you get a positive number. Since it cannot change signs on this interval, the derivative must be positive on all of $(-\infty,-1)$. That tells you that the original function, $f(x)$, is increasing on the interval $(-\infty,-1]$ (you get to include $-1$ as well).
To see what happens next, you should determine whether $f'(x)$ is positive or negative on $(-1,0)$; on $(0,1)$, and on $(1,\infty)$. This can be done by either using a "test point", or analyzing the derivative. For example, considering $(-1,0)$. Since $f'(x) =-4x(x^2-1)= -4x(x-1)(x+1)$, if we plug in a negative number for $x$ then $-4x$ will be positive. Because $x$ is between $-1$ and $0$, $x+1$ will be positive. And since $x$ is already negative, $x-1$ will be negative as well. So $f'(x) = -4x(x-1)(x+1)$ will be a product of two positive numbers and a negative number, so $f'(x)$ is negative on $(-1,0)$. That means that $f(x)$ is decreasing on $[-1,0]$.
Now do the same for the remaining intervals of constant sign for the derivative.