You start with f defined on the 1-dimensional subspace U, and you extend it one dimension at a time to the whole space. Vectors in U are of the form (–2y,y,0), and f(–2y,y,0) = –2y.
So as the first step, let's try to extend f to the subspace consisting of all vectors of the form (x,y,0). To do this, we have to assign a value α to f(1,0,0). Then (x,y,0) = (x+2y,0,0) + (–2y,y,0), and so .
We have to ensure that (because we have to keep the norm of the extension down to 2/3). You can convert that into a condition on α (just as in the proof of the H–B theorem), and I think you'll find that α has to have the unique value 2/3. Thus .
Now you have to extend f one dimension further, to the whole space. That means that you have to find a value β for f(0,0,1), without increasing the norm of f. I'll leave the details to you, but I think you'll find that β is not uniquely determined. In that way, you will be able to find different linear extensions of f with the same norm 2/3.