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,
$$x=a/b,\;\; y = b/a,\;\; z = \frac{(a+c)(b-c)}{ab}$$ 
 A: Consider the problem written as
\begin{equation*}
\frac{(1-f)^2}{f}+\frac{(1-g)^2}{g}+\frac{(1-h)^2}{h}+4=0
\end{equation*}
which gives the quadratic in $h$
\begin{equation*}
h^2+\frac{f^2g+f(g-1)^2+g}{f\,g}+1=0
\end{equation*}
As Tito stated, for this to have a rational solution (assuming $f,g$ rational) the discriminant must be a rational square. After simplification,
this means there must be $t \in \mathbb{Q}$ with
\begin{equation*}
t^2=f^2g^4+2f(f^2-1)^2g^3+(f^4-4f^3+4f^2-4f+1)g^2+2f(f-1)^2g+f^2
\end{equation*}
Defining $Y=t\,f$ and $X=g\,f$ gives the quartic
\begin{equation*}
Y^2=X^4+2(f-1)^2X^3+(f^4-4f^3+4f^2-4f+1)X^2+2(f^2-f)^2X+f^4
\end{equation*}
There is an obvious rational point $X=0, Y = f^2$, so the curve is birationally equivalent to an elliptic curve. Using Mordell's method
the elliptic curve is
\begin{equation*}
v^2=u^3+(f^4-4f^3-2f^2-4f+1)u^2+16f^4u
\end{equation*}
with
\begin{equation*}
g=\frac{u(f-1)^2-v}{2f(4f^2-u)}
\end{equation*}
The curve has a point of order $2$ at $u=0$, two points order $4$ when $u=4f^2$ and $4$ points of order $8$ at $u=4f$ and $u=4f^3$. These give undefined solutions
or permutations of $(f,-f,1)$ or $(f,-1/f,1)$. Thus the torsion subgroup is usually isomorphic to $\mathbb{Z}8$. There are $3$ rational points of order $2$ when
$f^2-6f+1=\square$ when the torsion subgroup is $\mathbb{Z}2 \times \mathbb{Z}8$.
Numerical tests show that the elliptic curve often has rank $0$, and so no other solutions. For $f=10$, the rank is $1$ with generator $(-125,-8250)$
which gives $g=-5/28$ and $h= -7/4,-4/7$.
If we set $f=(k^2-1)/2k$, the parametric solution quoted comes from
\begin{equation*}
u=\frac{(k-1)^2(k^6+2k^5-13k^4-4k^3-5k^2-6k-7)^2}{2(2k^6-4k^5+5k^4-12k^3+4k^2+1)^2}
\end{equation*} 
Numerical tests suggest that this is often $3$ times a generator point. Thus, it might be possible to find a smaller parametric solution by doing some further algebra.
A: 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.
