This replaces a message about Q1.55(b) that I posted earlier. My first answer was a bit too cryptic and incomplete. d denotes partial differentiation.
We have v(p1,p2,y)=w(p1,p2)+z(p1,p2)/y and p1>0, p2>0 and y>0. (Since we are dealing with zero demand if y=0 and p1>0 and p2>0, I feel justified in altering the question a little! It also removes the temptation to divide by zero.)
We need to find conditions on w and z to satisfy conditions 1 to 5 of Theorem 1.6 on page 28.
Condition 3 requires that v is strictly increasing in y, so z<0.
It is a bit difficult to make much further progress without assuming that both w and z are continuous and differentiable in (p1,p2) so we shall do so. Condition 1 is met by this assumption.
Condition 4 requires that v is decreasing in p1 and p2, so dw/dp1+dz/dp1.y<=0 (*). Letting y tend to zero gives dw/dp1<=0. Similarly dw/dp2<=0. Now divide(*)by y>0 and let y tend to infinity to get dz/p1<=0 and similarly dz/dp2<=0.
Condition 2 requires that v is homogeneous of degree zero in (p1,p2,y). So v(p1,p2,y)=v(tp1,tp2,ty) for all t>0. Write this out on full in w and z, differentiate with respect to t using the Chain Rule and the Quotient Rule, set t=1 and you should get the following expression (**):
0=dw/dp1.p1+dw/dp2.p2+{dz/dp1.p1+dz/dp2.p2-z}/y. Now let y tend to infinity to get that dw/dp1.p1+dw/dp2.p2=0. But condition 4 implied that dw/dpi<=0 for i=1,2 so dw/dpi=0 for i=1,2 as pi>0. Thus w(p1,p2)=k, a constant. (Check this!)
Now go back to (**), multiply through by y>0 and let y tend to zero to get dz.dp1.p1+dz/dp2.p2=z. Euler's Theorem (p.471) implies that z is homogeneous of degree 1 in (p1,p2).
Condition 5 implies that z is quasiconvex in (p1,p2). You can check there is no further restriction implied by v being quasiconvex in (p1,p2,y). (Remember that z<0).
I am happier with this answer than my earlier attempt.
Finally, you may think that Condition 1 has simply been assumed away and you are right! Although it is true that if f and g are continuous functions then f+g is also continuous, it is easy to think up functions f and g which are not continuous whereas f+g is continuous. For example, paste together two step functions. More exotically, consider f(x)=1 for x rational and f(x)=0 for irrational x together with g(x)=0 for x rational and g(x)=1 for irrational x. Neither f nor g is continuous anywhere but f+g=1. These are Dirichlet functions, if you want to look up the details.