To determine which of the given equations is definitely a quadratic equation in terms of $x$, we need to analyze each option:

Option A: $(\frac{1}{x})^2+\frac{2}{x}+1=0$

This equation involves terms with $x$ in the denominator, which means it is not a polynomial equation. A quadratic equation must be a polynomial of degree 2. Therefore, option A does not meet the requirements.

Option B: $x(x+1)=1+x^2$

Expanding and simplifying gives:
$$x^2 + x = 1 + x^2$$
Subtracting $x^2$ from both sides gives:
$$x = 1$$
This simplification leads to a linear equation, not a quadratic equation. Therefore, option B does not meet the requirements.

Option C: $x(x-1)=0$

Expanding gives:
$$x^2 - x = 0$$
This is a polynomial equation of degree 2, which means it is a quadratic equation. Therefore, option C meets the requirements.

Option D: $ax^2 + x + 1 = 0$

This equation is a quadratic equation when $a \neq 0$. However, if $a = 0$, the equation becomes $x + 1 = 0$, which is a linear equation. Since the condition "definitely a quadratic equation" implies it must always be quadratic regardless of the coefficients, option D does not meet the requirements when $a = 0$.

Based on the analysis, the only option that is definitely a quadratic equation in terms of $x$ is:

$$\boxed{C}$$