Solve the system$$\left\{\begin{array}{l}3b-a=x\\3a+c=y\\3b-c=z.\end{array}\right.$$You will get$$a=\frac{1}{4} (-x+y+z),\ b=\frac1{12}(3x+y+z)\text{ and }c=\frac14(3x+y-3 z).$$So,$$f^{-1}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}\frac{1}{4} (-a+b+c)\\\frac1{12}(3a+b+c)\\\frac14(3a+b-3c)\end{pmatrix}.$$
Your equation is
$$\begin{align*}
\frac{\sqrt{y}}{y-1} = x \\
\implies\\
\frac{y}{(y-1)^2} = x^2 \\
\iff\\
x^2y^2 -(2x^2 + 1)y + x^2 = 0
\end{align*}$$
This is a second degree polynomial in the variable $y$, so you get two roots ($\Delta = (2x^2+1)^2 - (2x^2)^2 = 4x^2+1$).
$$y_1 = \frac{2x^2 + 1 + \sqrt{\Delta}}{2x^2} \qquad \text{or} \qquad y_2 = \frac{2x^2 + 1 - \sqrt{\Delta}}{2x^2}$$
You see that you have two "valid" inverses (both are positive), but why is it possible? Your fonction is not defined at $1$, it is only defined in
$$[0,1[ \cup ]1,\infty[$$
But it first goes in the negative (before 1) then in the positive (after 1). If you've read this carefully, you've noticed that there is an implication (this is why you think you have two inverses) which goes from giving information from $x$ to information from $x^2$. This means we still need to do the backward step.
Since $y_1$ is always strictly positive (and greater than $1$) we can easily deduce that the right inverse for the negative is $y_2$ and the right inverse for the (strictly) positive is $y_1$!
In fact, it is bijective continuous from $\mathbb{R}_+\setminus \{1\}$ to $\mathbb{R}$and the inverse (which is unfortunately not continuous but continuous by parts) is given by $f^{-1}(x) = \frac{2x^2 + 1 + s(x) \sqrt{4x^2+1}}{2x^2}$ where $s(x) = -1$ if $x \leqslant 0$ and $s(x) = 1$ if $x >0$. It is defined by continuity on the left in $0$ but if you wish it to be clearer $f^{-1}(0) = 0$.
We deduce $f^{-1}(1) = \frac{3 + \sqrt{5}}{2}$.
For further comprehension, check the graph on any calculator/computer e.g. Wolfram.
I also suggest to draw variation tables to have a proper idea! (study the derivative).
Best Answer
In general, finding a formula for the inverse of a 2-var function can be very very hard. For the specific case of a function like this one ("linear in each variable") we can do it with basic algebra. Write \begin{align} u &= ax + by\\ v &= cx + dy. \end{align} The goal is then to find expressions for $x$ and $y$ just in terms of $u$ and $v$.
Multiply the top equation by $d$ and the bottom by $b$ to make the $y$ terms the same: \begin{align} du &= adx + bdy\\ bv &= bcx + bdy. \end{align} Subtract the top equation from the bottom to get \begin{align} du &= adx + bdy\\ bv - du &= bcx - adx = (bc - ad)x. \end{align} And now you can solve that last equation for $x$, getting \begin{align} du &= adx + bdy\\ \frac{bv - du}{bc - ad} &= x. \end{align} You can either plug that into the first equation and solve for $y$, or you can return to the start and multiply the top equation by $c$ and the bottom by $a$ to get \begin{align} cu &= acx + bcy\\ av &= acx + ady. \end{align} Subtract to get \begin{align} cu &= acx + bcy\\ av - cu &= ady - bcy = (ad - bc)y \end{align} and then divide the last equation to get \begin{align} \frac{av - cu}{ad - bc} &= y \end{align}
and there are your two formula!