I'm not crazy about the wording. "Functions" are not geometric objects so there can't be an area between them! What is meant is the area between the graphs of the two functions.
Yes, the first thing I would do is look at the graphs of the equations. The second thing I would do is determine where the graphs intersect. That will be where \(\displaystyle y= \frac{x^3}{3}- x= \frac{x}{3}\). Since I hate fractions I would immediately multiply both sides by "3": \(\displaystyle x^3- 3x= x\) so \(\displaystyle x^3- 4x= x(x^2- 4)= x(x- 2)(x+ 2)= 0\). A product is 0 if and only if at least one of the terms is 0 so we must have x= 0, x- 2= 0, or x+ 2= 0. That is, x= 0, x= 2, and x= -2. Those look right on your graph, don't they?
The third thing I would do is notice that, for x between -2 and 0, the straight line graph, the graph of \(\displaystyle y= \frac{x}{3}\), is below the curved graph, the graph of \(\displaystyle \frac{x^3}{3}- x\), and for x between 0 and -2 it is the opposite. Since area is always positive we need to integrate a positive function so, for x from -2 to 0 we need to integrate \(\displaystyle \frac{x^3}{3}- x- \left(\frac{x}{3}\right)\) and for x from 0 to 2 we need to integrate \(\displaystyle \frac{x}{3}- \left(\frac{x^3}{3}- x\right)\).
The area betw0een the two graphs is
\(\displaystyle \int_{-2}^0 \frac{x^3}{3}- x- \frac{x}{3}dx+ \int_0^2 \frac{x}{3}- \frac{x^3}{3}+ x dx\)
\(\displaystyle = \int_{-2}^0 \frac{x^3}{3}- \frac{4x}{3}dx+ \int_0^2 \frac{4x}{3}- \frac{x^3}{3} dx\).
