Problem II 4.1(b)  Find the integral surface for
V = (x, -y, 0) containing the curve C: x = t, y = t, z = t^2; 0 < t < infinity.
The equation for the integral surface is wrong.  You need to figure it out and run it to show that the curve lies in the surface.

u1[x_,y_,z_] := x y
u2[z_, z_, z_] := z

curve1 = ParametricPlot3D[{t, t, t^2}, {t, 0, 1}];
surface1 = ContourPlot3D[x y -z  == 0, {x, 0,1}, {y, 0, 1}, {z, 0, 1}];
Show[curve1, surface1, AxesLabel ->{x,y,z}]

Well, that was sort of boring.  How about the integral surface for the same vector field that contains the helix
x = cos(t), y = sin(t), z = t/π  for all t.
U1(t) = cos(t) sin(t) = 1/2 sin(2t)
U2(t) = t/Pi.
solve second for t = Pi U2(t)/
Thus U1(t) - 1/2 sin(2 (Pi U2(t)) = 0, so
u1(x,y,z) - 1/2 sin(2 Pi u2(x,y,z)) = 0. Using substituting in the functions u1 and u2 gives
x y - 1/2 sin(2 Pi z) = 0.

curve2 = ParametricPlot3D[{Cos[t], Sin[t], t/π}, {t, -π, π}];
surface2 = ContourPlot3D[x y -1/2 Sin[2 Pi z]  == 0, {x, -1.5,1.5}, {y, -1.5, 1.5}, {z, -1, 1}];
Show[curve2, surface2, AxesLabel ->{x,y,z}]

In general, this is just a local result.
Suppose we want an integral surface in some neightborhood of (1, 0, 0), which is the point on C at t = 0.

We can solve U1(t) = 1/2 sin(2t) for t:
t = 1/2 arcsin(2 U1(t)), which holds for -1/2 < U1(t) < 1/2
So 
U2(t) = U2(1/2 arcsin(2 U1(t)), or
U2(t) - U2(1/2 arcsin(2 U1(t)) = 0.
Thus
u2(x,y,z) - u2(1/2 arcsin(2 u1(x,y,z)) = 0,
which holds for -1/2 < u1(x,y,z) < 1/2.
Plugging in u1 and u2 gives
z - 1/Pi (1/2 arcsin(2 (x y))) = 0, which simplifies to 
z - 1/(2Pi) arcsin(2 x y) = 0
which holds for -1/2 < x y < 1/2

curve2 = ParametricPlot3D[{Cos[t], Sin[t], t/π}, {t, -π, π}];
surface2 = ContourPlot3D[z -1/(2Pi) ArcSin[2 x y]  == 0, {x, -1.5,1.5}, {y, -1.5, 1.5}, {z, -1, 1}];
Show[curve2, surface2, AxesLabel ->{x,y,z}]