Ok, it made it easier for me to think about but forget about what i said about cutting it in half.
If x were greater than 36, then you would be outside the ellipse, but you want to say inside.
Think of creating your rectangle using the variable x. If I set one corner at (x,0), then I can set another corner at (x, sqrt(1296-x^2)). So it has width 2x and height sqrt(1296-x^2). Therefore the area is ____ ?
Regarding the 75% thing - for what x will the height be exactly 75% of the width? Now for what values is it less than 75% of the width? Think that you want sqrt(1296-x^2) to be less than 75% of 2x.