So, I suppose that you have three data points ($i=1,2,3)$ $$y_i=\frac{a}{b+(x_i-c)^2}$$ from which you want to extract parameters $a,b,c$.
If you compute the ratio $A_{12}=\frac{y_2}{y_1}$, $a$ is gone and you can extract $b$. you can do the same with $A_{13}=\frac{y_3}{y_1}$ and extract $b$ again. Write that these two values of $b$ are equal and you obtain a linear equation in $c$ which is easy to solve. Now, go backwards to get successively $b$ and $a$. After simplifications, you should then get
$$c=\frac{1}{2}\frac{{x_1}^2 {y_1} ({y_2}-{y_3})+{x_2}^2 {y_2}
({y_3}-{y_1})+{x_3}^2 {y_3} ({y_1}-{y_2})}{ {x_1}
{y_1} ({y_2}-{y_3})+{x_2} {y_2}
({y_3}-{y_1})+{x_3} {y_3} ({y_1}-{y_2})}$$ $$b=\frac{{y_2} (c-{x_2})^2-{y_1} (c-{x_1})^2}{{y_1}-{y_2}}$$ $$a={y_1} \left(b+(c-{x_1})^2\right)$$
Added later
Since this is to be used for regression and since this model is nonlinear with respect to its parameters, linearization would provide reasonable starting guesses for the nonlinear regression.
So, let us start with $$y=\frac{a}{b+(x-c)^2}$$ and rewrite it as $$\frac{1}{y}=\frac{x^2-2cx+(b+c^2)}{a}=\alpha x^2+\beta x+\gamma$$ The quadratic regression based on points$(x_i,\frac{1}{y_i})$ will give $\alpha,\beta,\gamma$. So, now $$a=\frac{1}{\alpha},b=\frac{4 \alpha \gamma -\beta ^2}{4 \alpha ^2},c=-\frac{\beta }{2 \alpha }$$
But again, the values are just starting guesses since in the linearized model you minimize $$SSQ_1=\sum_{i=1}^N \Big(\frac{1}{y_i^{exp}}-\frac{1}{y_i^{calc}}\Big)^2$$ while, since the $y_i$ are the measured values, you must minimize $$SSQ_2=\sum_{i=1}^N \Big({y_i^{exp}}-{y_i^{calc}}\Big)^2$$
Illustrating example
To illustrate this aspect, I generated data using $a=123.4,b=5.6,c=7.8$ for $x$ varying between $0$ and $20$ ($21$ data points) and the exact values of $y$ were randomly perturbed (relative change between $-10$% and $+10$%).
The linearization step leads to $\alpha=8.323492415\times 10^{-3}$,$\beta=-1.299834856\times 10^{-1}$,$\gamma=5.468572881\times 10^{-1}$ to which correspond $a=120.142$, $b=4.732$, $c=7.808$. These do not seem bad when compared to the values used for the generation of the data. However, for these values, the sum of squares is $45.118$ and the nonlinear regression ends with a sum of squares equal to $3.932$ that is to say almost $12$ times smaller and the final parameters are $a=125.559$, $b=5.998$, $c=7.799$. Big differences !
Looks correct, but you can always check this yourself by plugging your $u_e$ into the expression for either $\dot{x}_2$ or $\dot{y}_2$.
Also note that the input only appears as $u^2$, so using $-u$ should have the same effect as $u$.
Lastly it is also possible to solve for $u_e$ by setting $\dot{x}_2=0$ while using $x_1=h_0$. However, you might also have to linearize the system around this equilibrium point, so doing this coordinate translation would probably have to be done anyway in a later stage.
Best Answer
It is ok. When you transform $x\to y$ your old equilibrium $(1,0)$ becomes $(0,0)$ for the $y$-system. You only have to analyze that point for the new system. You cannot avoid by translation having several equilibria.