The parameters enter nonlinearly into your equation. However, fortunately you can rewrite
your equation in such a way that you have linear parameters:
$$ (f(x+1) - f(x))^{1/3} = a f(x) - b $$
where $k_1 = a^3$ and $k_2 = b/a$.
So you can apply least squares to these linear equations in $a$ and $b$ for $x$ from $1$ to $n$.
EDIT: I tried some data that seem close to what you had in your picture:
$$ y = [237,130,120, 113, 111, 110] $$
Linear least squares for the residuals $(y_{i+1} - y_i)^{1/3} - a y_i + b$, $i = 1 \ldots,5$, in Maple produces
$$ a = -.0273525754168209,\ b = -1.67458695987193$$
which corresponds to
$$ k_1 = - 0.0000204641953284227606,\ k_2= 61.2222774036158200$$
I then tried nonlinear least squares for the residuals
$ y_{i+1} - y_i - k_1 (y_i - k_2)^3$, $i=1 \ldots 5$, using the above as initial values
(this is often necessary because the nonlinear least squares algorithms often provide only a local minimum rather than the global minimum). The result was
$$ k_1 =- 0.0000166852730695581124, k_2 = 51.1767746591434971 $$
The recursion using initial value $237$ and parameters from the linear least squares
produces the values
$$ 237, 125.855961638565, 120.330463855443, 116.104385196179, 112.721501867265, 109.926405798693$$
while the recursion using parameters from the nonlinear least squares produces
$$ 237, 129.938505804843, 121.786225982822, 115.912389815574, 111.385883230420, 107.744051206183$$
For an easier solution, consider that you have $n$ data points $(x_i,y_i)$ and you look for the equation of the circle. So, ideally, $$f_i=(x_i-x_c)^2+(y_i-y_c)^2-r^2$$ Now compute $f_j-f_i$ $$F_{i,j}=f_j-f_i=2(x_i-x_j)x_c+2(y_i-y_j)y_c+(x_j^2+y_j^2)-(x_i^2+y_i^2)$$ which is the form of a linear regression. From it, you will easily compute $(x_c,y_c)$ by a linear regression with no intercept.
Another way is to consider the general equation of conics $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ For a circle $A=C$ and $B=0$ which makes the equation to be $$x^2+y^2+\alpha x+\beta y+\gamma=0$$ Again, a linear regression will give you $\alpha,\beta ,\gamma$ and then the classical transform will provide $x_c,y_c,r$.
At least, these will be good estimates for starting the minimization of any objective function of your choice.
Best Answer
Here is a very simple-minded way.
First, set $V = \max(|y_i|)$.
Then, let $\phi_i =\arcsin(y_i/V)-x_i $.
Finally, $\phi =\frac1{n}\sum \phi_i $.
Note: If the data is over multiple sinusodial cycles, adjust $\phi_{i+1}$ so it differs from $\phi_i$ by less than $\pi$.