Solutions of : $y' = (y^2 + z^2 +1)^{-a} , z' = y(1+z^2)^a$ 
Study the existence of solutions that are set entirely on $\mathbb R$ for the functions :
$$  y' = (y^2 + z^2 +1)^{-a}, z' = y(1+z^2)^a  $$

I came upon this problem while studying for my Dynamical Systems course, but I'm not sure on how to proceed.
One thing I saw was that we could bound the second equation, such as :
$$z' = (1 +z^2)^a \leq  (1+y^2 + z^2)^a = y/y'$$
since $y^2 \geq \forall y\in C(\mathbb R).$
So, we have that the second equation is bound between the solution and the derivative of the first equation, since :
$$z \leq \frac{y}{y'}$$
but I cannot see how to use this in order to study the existence of the equation (it can maybe help on proving Lipschitz conditions for the uniqueness but that's not what I need here).

Of course I don't know if the following considerations are helpful for your course but at least it can be seen that a solution for $\,y\,$ and $\,z\,$ exists. 
With $\enspace\displaystyle y=\frac{z'}{(1+z^2)^a}\enspace $ follows 
$\enspace\displaystyle y'= \frac{z''}{(1+z^2)^a}-  \frac{2az(z')^2}{(1+z^2)^{a+1}} \,$ . 
With substituting $\,y'\,$ and $\,y\,$ in $\enspace y'(y^2+z^2+1)=1\enspace$ and the equation solved for $\,z''\,$ we get:
$$ z''= \frac{2az(z')^2}{1+z^2} +\frac{(1+z^2)^{3a}}{((z')^2+(1+z^2)^{2a+1})^a} $$ 
We see now, that one gets the $\,\text{n}^{th}$ derivation $\,z^{(n)}$ 
by $\,z^{(k+1)}=(z^{(k)})'\,$ in dependence of $\,z\,$ and $\,z'\,$ .
The denominators are always positive (in $\mathbb{R}$) and therefore exists a Taylor series as a solution 
for $\,z\,$ . $\,\,$(And $\,y\,$ can be calculated by $\,z\,$ and $\,z'\,$ because of the initial formula above.)
Example for the first values:   
$z(0):=0\,$ , $\,z’(0):=0\enspace$  =>  $\enspace z’’(0)=1\,$ ; $\,\,y(0)=0\,$ , $\,y’(0)=1$  
