<s>[INST] What would the space under the image of this function look like? The function is as follows: $$\phi(x,y,z) = \left(\frac{yz}{x^2+y^2+z^2},\frac{zx}{x^2+y^2+z^2}\right)$$
And when $z=0$,the function becomes $ \; \phi(x,y,0)=\frac{xy}{x^2+y^2}$. Some properties of this function are that for any point $ \; p = (x,y,z) \in \mathbb{R}^3/(0) \; and \; \forall \lambda \in \mathbb{R} \neq 0$; if $\; \lambda p=(\lambda x,\lambda y,\lambda z)$ then 
$$\begin{align}
\phi(\lambda p) & =\left(\frac{\lambda y \lambda z}{ \lambda^2 x^2 + \lambda^2 y^2 + \lambda z^2},
\frac{\lambda z \lambda z}{ \lambda^2 x^2 + \lambda^2 y^2 + \lambda z^2}\right)\\
& = \left(\frac{\lambda^2(yz)}{\lambda^2(x^2+y^2+z^2)},\frac{\lambda^2(zx)}{\lambda^2(x^2+y^2+z^2)}\right) \\
& = \phi(p)\end{align}$$ and so of course $\phi(-x,-y,-z)=\phi(x,y,z)$. I'm tempted to say this surface is going to look an awful lot like the projective plane and share certain features of it (like mapping 2 antipodal points into point under the image of $\phi$) but I can't exactly visualize it. 
 [/INST] I claim that the image equals the closed ball $B$ with radius $1/2$ and center $(0,0)$.
For $\mathrm{im}(\phi) \subseteq B$, observe that
$$\left(\frac{yz}{x^2+y^2+z^2}\right)^2 + \left(\frac{zx}{x^2+y^2+z^2}\right)^2=\frac{z^2 (y^2+x^2)}{(x^2+y^2+z^2)^2} \leq \frac{1}{4}$$
since for all $a,b,c \geq 0$ we have $(a+b+c)^2-4c(a+b)=(a+b-c)^2 \geq 0$.
I leave the other inclusion as an exercise.
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