Proof of same two-parameter family curves in $\mathbb R^3$. Exercise :
If $V = (x,y,z)$ and $\Omega = \{(x,y,z) \in \mathbb R^3 : x>0, y>0, z>0\}$, show that $yz/x^2$ and $(y^2-z^2)/x^2$ are functionally independent first integrals of $V$ in $\Omega$. Verify directly that the equations 
$$\frac{yz}{x^2} = c_1, \quad \frac{(y^2-z^2)}{x^2}=c_2$$
and the equations of the example (1.21)
$$\frac{y}{x} = c_1, \quad \frac{z}{x} = c_2$$
describe the same two-parameter family of curves in $\mathbb R^3$.
Attempt :
Let : $$u_1 = \frac{yz}{x^2}, \quad u_2 = \frac{y^2-z^2}{x^2}$$
The functionals $u_1, u_2$ indeed are first integrals of the vector field $V$, since it is $u_1, u_2 \in C^1(\Omega)$ and also 
$$xu_{1_x} + yu_{1_y} + zu_{1_z} = x\frac{-2yz}{x^3} + y\frac{z}{x^2} + z \frac{y}{x^2} = 0$$
$$xu_{2_x} + yu_{2_y} + zu_{2_z} = x \frac{-2(y^2-z^2)}{x^3} + y\frac{2y}{x^2} + z \frac{-2z}{x^2}= 0 $$
Finally, we have that
$$\nabla u_1 = \bigg(-\frac{2yz}{x^3}, \frac{z}{x^2}, \frac{y}{x^2}\bigg), \quad \nabla u_2 = \bigg(-\frac{2(y^2-z^2)}{x^3}, \frac{2y}{x^2}, -\frac{2z}{x^2}\bigg) $$
$$\text{and}$$
$$\nabla u_1 \times \nabla u_2 = \begin{vmatrix} i & j & k \\ -2yz/x^3 & z/x^2 & y/x^2 \\ -2(y^2-z^2)/x^3 & 2y/x^2 & -2z/x^2 \end{vmatrix} = \dots = 0$$
from where we conclude the linear independence.
Question : Now, for the last part regarding proving that the equations describe the same two parameter family of curves in $\mathbb R^3$, is where I am at a loss at continuing. I can observe that, the second given equations from the example (1.21) yield the first ones if you multiply them and if you take the difference of their squares, but how would I conclude that they describe the same two-parameter family of curves ?
 A: Assume that we have two functionally independent first integrals, $u_1$ and $u_2$.  At each point, the gradient of any $C^1$ function is orthogonal to the level set of the function passing through that point.  The condition for the gradients of both first integrals having nonzero cross-product is equivalent to their gradients being linearly independent, in other words, not collinear.  The level sets of $u_1$ and $u_2$, being orthogonal to their respective gradients, intersect along a $C^1$ curve (this follows, e.g., from the constant rank theorem).  The level sets of a first integral are mutually disjoint and cover the whole of $\Omega$, hence $\Omega$ can be represented as the disjoint union of sets of the form
$$
\{(x,y,z) \in \Omega : u_1(x,y,z) = c_1\} \cap \{(x,y,z) \in \Omega : u_2(x,y,z) = c_2\},
$$
where $c_i$'s belong to the range of the function $u_i$, $i = 1, 2$.
In your case, fix $c_1$ and $c_2$ and consider the curve
$$
\{(x,y,z) \in \Omega : \frac{y}{x} = c_1, \frac{z}{x} = c_2 \}.
$$
Then, for any $(x, y, z)$ belonging to the above set we have
$$
\frac{yz}{x^2} = c_1 c_2 =: \tilde{c}_1, \quad \frac{y^2 - z^2}{x^2} = (c_1)^2 - (c_2)^2 = : \tilde{c}_2.
$$
Consequently, any intersection of level sets of the functions $\frac{y}{x}$ and $\frac{z}{x}$ is contained in the intersection of some level sets of the functions $\frac{yz}{x^2}$ and $\frac{y^2 - z^2}{x^2}$.  Similarly one can show that any intersection of level sets of the functions $\frac{yz}{x^2}$ and $\frac{y^2 - z^2}{x^2}$ is contained in the intersection of some level sets of the functions $\frac{y}{x}$ and $\frac{z}{x}$.
