Can this equation be solved in polynomial time?

Question

I came across a more general form of this question.

Can we find the value of variables in polynomial time ?

Let $m = n^{2}$, there are $m$ variables ($x,y,z\ldots$) in the equation and these $m$ values can only take positive integer values ranging from 1 to $n$.

We have equation generator,
$$a\cdot f(x)+b\cdot f(y)+c\cdot f(z)+\cdots=k$$

for variables $x,y,z\in \{1,2\ldots, n\}$.

$f$ can be any thing such as $x, x^{2}, \log x$ etc.

Given that the constants are rational numbers and that there exists a solution, can we solve for the variables in polynomial time?

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Update: Please note that I am not giving a single instance of $a\cdot f(x)+b\cdot f(y)+c\cdot f(z)+\cdots=k$, instead I am giving you the entire family of all possibilities of the above equation.

Basically I am giving you a black box, if you input $f$ it gives you corresponding $k$, given that $f$ is computable in polynomial time. We can do this multiple times. The goal is to solve for variables using this black box, in a polynomial time.

So we have:
1. The values of $m$ and $n$
2. The constants $a,b,c ...$
3. The equation generator $a\cdot f(x)+b\cdot f(y)+c\cdot f(z)+\cdots=k$

We can go on choosing different $f$ and the generator will give us corresponding equation, with a new $k$.

Goal: To solve for variables in the equation when $f(x) = x$

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Update 2: For people wondering the origin of the question and its purpose. If the answer for this question is yes, then I think Sudoku can be solved in polynomial time. I tried to make use of the numerical property of Sudoku, that the sum of the entries of any row or column is known. By dealing each unknown cell as a variable we could treat each row and column as an equation. For a Sudoku of $n*n$ it would result in $2n$ equations with $n^2-k$ variables where $k$ is the number of already filled cells. This is an underdetermined system of equations. We could represent this in a matrix and it can be reduced to Echelon form. This can be repeated for sum of squares, cubes etc etc of the values. If we could solve this we could solve Sudoku.