Let . Then f is identically zero on the boundary of the unit cube, so its maximum must occur at a critical point inside the cube. So put the three partial derivatives equal to 0. Leaving out some details, this gives the equations
From (1) and (2) it's easy to see that x=y. From equations (1) and (3) (with y=x) you see that . Substitute that into (1) and solve for x, to get .
Finally, check that . So that is the maximum value of the function in the cube.