x^3+y^3=64.
Values that satisfy this equation are:
x=0, y=4 since 0^3+4^3=4^3=4x4x4=16x4=64.
Similarly, x=4, y=0 since 4^3+0^3=64.
Thus there will be points (0,4), and (4,0) on the graph.
To determine how the rest of the graph looks like, we can put it in a form of y=(-1)^(2/3)cuberoot(64-x^3) by solving for y. Then one may take the 1st/2nd derivatives to find where the function increases and decreases or where it is concave up and down. I'll leave those details for you to work out.
To see how the graph looks like, go to www.wolframalpha.com and in the search bar type in "x^3+y^3=64" and a graph of this function will appear.
