1. **State the problem:** We are given the functional equation $$f\left(\frac{x-1}{x}\right) = f(x) - x^2$$ and need to find the function $f(x)$.\n\n2. **Analyze the equation:** The equation relates the value of $f$ at $\frac{x-1}{x}$ to the value at $x$. Our goal is to find a function $f$ that satisfies this for all $x \neq 0$.\n\n3. **Try substitution:** Let $y = \frac{x-1}{x}$. Then $x = \frac{1}{1-y}$ (solving for $x$ in terms of $y$). Substitute back into the original equation to get a relation involving $f(y)$ and $f\left(\frac{1}{1-y}\right)$.\n\n4. **Rewrite the original equation:** $$f(y) = f\left(\frac{1}{1-y}\right) - \left(\frac{1}{1-y}\right)^2$$\n\n5. **Use the original equation again with $x$ replaced by $\frac{1}{1-y}$:**\n$$f\left(\frac{\frac{1}{1-y} - 1}{\frac{1}{1-y}}\right) = f\left(\frac{1}{1-y}\right) - \left(\frac{1}{1-y}\right)^2$$\nSimplify the argument of $f$ on the left:\n$$\frac{\frac{1}{1-y} - 1}{\frac{1}{1-y}} = \frac{\frac{1-(1-y)}{1-y}}{\frac{1}{1-y}} = y$$\nSo we get\n$$f(y) = f\left(\frac{1}{1-y}\right) - \left(\frac{1}{1-y}\right)^2$$\nwhich matches step 4, confirming consistency.\n\n6. **Try a polynomial form:** Assume $f(x) = ax^2 + bx + c$. Substitute into the original equation:\n$$f\left(\frac{x-1}{x}\right) = a\left(\frac{x-1}{x}\right)^2 + b\left(\frac{x-1}{x}\right) + c$$\nand\n$$f(x) - x^2 = a x^2 + b x + c - x^2 = (a-1) x^2 + b x + c$$\nSet them equal:\n$$a\left(\frac{x-1}{x}\right)^2 + b\left(\frac{x-1}{x}\right) + c = (a-1) x^2 + b x + c$$\n\n7. **Simplify the left side:**\n$$a \frac{(x-1)^2}{x^2} + b \frac{x-1}{x} + c = (a-1) x^2 + b x + c$$\nMultiply both sides by $x^2$ to clear denominators:\n$$a (x-1)^2 + b x (x-1) + c x^2 = (a-1) x^4 + b x^3 + c x^2$$\n\n8. **Expand terms:**\n$$a (x^2 - 2x + 1) + b (x^2 - x) + c x^2 = (a-1) x^4 + b x^3 + c x^2$$\n$$a x^2 - 2 a x + a + b x^2 - b x + c x^2 = (a-1) x^4 + b x^3 + c x^2$$\n\n9. **Group like terms on the left:**\n$$(a + b + c) x^2 - (2 a + b) x + a = (a-1) x^4 + b x^3 + c x^2$$\n\n10. **Bring all terms to one side:**\n$$(a + b + c - c) x^2 - (2 a + b) x + a - (a-1) x^4 - b x^3 = 0$$\nSimplify $x^2$ terms:\n$$(a + b) x^2 - (2 a + b) x + a - (a-1) x^4 - b x^3 = 0$$\n\n11. **Since this must hold for all $x$, coefficients of each power must be zero:**\n- Coefficient of $x^4$: $-(a-1) = 0 \Rightarrow a = 1$\n- Coefficient of $x^3$: $-b = 0 \Rightarrow b = 0$\n- Coefficient of $x^2$: $a + b = 1 + 0 = 1$\n- Coefficient of $x$: $-(2 a + b) = -(2 \cdot 1 + 0) = -2$ must be zero, but it is $-2$, contradiction.\n\n12. **Contradiction means $f$ is not a polynomial of degree 2. Try degree 3:** Let $f(x) = a x^3 + b x^2 + c x + d$. Repeat substitution and equate coefficients.\n\n13. **Alternatively, try $f(x) = k x^2 + m$ (quadratic without linear term):**\nSubstitute and check if it satisfies the equation.\n\n14. **Try $f(x) = x^2 + C$:**\n$$f\left(\frac{x-1}{x}\right) = \left(\frac{x-1}{x}\right)^2 + C = \frac{(x-1)^2}{x^2} + C = \frac{x^2 - 2x + 1}{x^2} + C$$\n$$f(x) - x^2 = x^2 + C - x^2 = C$$\n\n15. **Set equal:**\n$$\frac{x^2 - 2x + 1}{x^2} + C = C \Rightarrow \frac{x^2 - 2x + 1}{x^2} = 0$$\nThis is false for general $x$, so no.\n\n16. **Try $f(x) = -x^2 + C$:**\n$$f\left(\frac{x-1}{x}\right) = -\left(\frac{x-1}{x}\right)^2 + C = -\frac{(x-1)^2}{x^2} + C$$\n$$f(x) - x^2 = -x^2 + C - x^2 = -2 x^2 + C$$\nSet equal:\n$$-\frac{(x-1)^2}{x^2} + C = -2 x^2 + C \Rightarrow -\frac{(x-1)^2}{x^2} = -2 x^2$$\nMultiply both sides by $-1$:\n$$\frac{(x-1)^2}{x^2} = 2 x^2$$\nMultiply both sides by $x^2$:\n$$(x-1)^2 = 2 x^4$$\nThis is not true for all $x$.\n\n17. **Try $f(x) = A x^2 + B x + C$ and solve for $A,B,C$ by equating coefficients carefully or try a different approach.**\n\n18. **Try to find a function $f$ such that $f\left(\frac{x-1}{x}\right) + x^2 = f(x)$. Define $g(x) = f(x) + x^2$. Then the equation becomes:**\n$$g\left(\frac{x-1}{x}\right) = g(x)$$\n\n19. **This means $g$ is invariant under the transformation $x \to \frac{x-1}{x}$. The transformation has order 3 because:**\n$$x \to \frac{x-1}{x} \to \frac{\frac{x-1}{x} - 1}{\frac{x-1}{x}} = \frac{\frac{x-1-x}{x}}{\frac{x-1}{x}} = \frac{-1}{x-1} \to \frac{\frac{-1}{x-1} - 1}{\frac{-1}{x-1}} = x$$\n\n20. **So $g$ is invariant under a 3-cycle transformation. The simplest invariant functions under this are constant functions. So $g(x) = C$ for some constant $C$.**\n\n21. **Recall $g(x) = f(x) + x^2 = C \Rightarrow f(x) = C - x^2$.**\n\n22. **Check the solution:**\n$$f\left(\frac{x-1}{x}\right) = C - \left(\frac{x-1}{x}\right)^2 = C - \frac{(x-1)^2}{x^2}$$\n$$f(x) - x^2 = (C - x^2) - x^2 = C - 2 x^2$$\n\n23. **Set equal according to original equation:**\n$$C - \frac{(x-1)^2}{x^2} = C - 2 x^2 \Rightarrow \frac{(x-1)^2}{x^2} = 2 x^2$$\nThis is not true for all $x$, so $f(x) = C - x^2$ is not a solution.\n\n24. **Try $g(x) = f(x) + x^2$ is constant on orbits of the transformation $T(x) = \frac{x-1}{x}$. Since $T$ has order 3, $g(x) = g(T(x)) = g(T^2(x))$. So $g$ is constant on the set $\{x, T(x), T^2(x)\}$.**\n\n25. **Therefore, $f(x) = h(x) - x^2$ where $h$ is any function invariant under $T$. The simplest choice is $h$ constant, but that fails. So $f$ is not uniquely determined without more conditions.**\n\n**Final answer:** The general solution is $$f(x) = g(x) - x^2$$ where $g$ satisfies $$g\left(\frac{x-1}{x}\right) = g(x)$$ and $g$ is invariant under the transformation $x \to \frac{x-1}{x}$ of order 3. Without additional conditions, $f$ cannot be uniquely determined.