On the equation $(1-x)^2/x + (1-y)^2/y + (1-z)^2/z + 4 = 0$ The problem is to solve the equation,
$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$
in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would be to make the discriminant $D$ of $(1)$ a square,
$$D = -4x^2y^2 + (x+y-2xy+x^2y+xy^2)^2 = t^2\tag{2}$$
with one solution being,
$$ x = a/b$$
$$ y = -b/a\, (p_1/p_2)$$
$$p_1 = 7a^6 + 4a^5b - 14a^4b^2 + 12a^3b^3 - 25a^2b^4 + 8a b^5 - 8b^6$$
$$p_2 = 8a^6 - 8a^5b + 25a^4b^2 - 12a^3b^3 + 14a^2b^4 - 4a b^5 - 7b^6$$
and $a,b$ being the legs of the Pythagorean triple $a^2+b^2 = c^2$.
However, can someone find a polynomial parameterization of small degree to $(1)$?
Edit (a few days later):
Courtesy of Allan MacLeod, a simple parameterization to $(1)$ can be given by using $x=1/y$ and the discriminant $(2)$ greatly simplifies to just solving,
$$y^2+1 = w^2$$
Hence, his first answer below can also be expressed in terms of Pythagorean triples $a^2+b^2 = c^2$ as,
$$x=a/b,\;\; y = b/a,\;\; z = \frac{(a+c)(b-c)}{ab}$$ 

After some further thought and computation, if $f=(k^2-1)/2k$ then $u=2(k-1)^2$ gives
\begin{equation*}
v= \pm \frac{(k-1)^2(k^2-2k-1)(k^2-2k+3)}{2k^2}
\end{equation*}
This point gives the parametric form
\begin{equation*}
f=\frac{k^2-1}{2k} \hspace{2cm}
g=\frac{2k}{k^2-1} \hspace{2cm}  h=\frac{k(1-k)}{k+1}
\end{equation*}
Doubling this point (with $f=(k^2-1)/2k$) gives
\begin{equation*}
g=\frac{(1-k)(k^2+2k-1)(3k^2+2k+1)}{2k(k+1)(k^2+1)(k^2-2k+3)}
\end{equation*}
\begin{equation*}
h=\frac{k(k+1)(k^2+2k-1)(k^2-2k+3)}{2(k-1)(k^2+1)(3k^2+2k+1)}
\end{equation*}
Further forms can probably be derived using the torsion points.
