Your question is obviously related to math.stackexchange.com/questions/1214044/…, please reference this. In my opinion, both your answer there and your question here are incomprehensible.

You haven't said a word about what $f(x)$ is. The number of common points between the curves $f(x)^3$ and $x^2$ is infinite when $f(x) = x^{2/3}$.

Sorry for inconvenience. $y=f(x)$ such that $\frac{d}{dx}x^{2} \neq\frac{d}{dx} f(x)^3$, this is given.

certainly , $f(x)\neq x^{2/3}$ since $f(x)=x^{2/3}$violates the given condition.

Your stated reasoning is incorrect: integrating both sides of a $\ne$ inequality does not yield an inequality. You need an assumptions like $f(x)$ being everywhere differentiable, then this follows (in the one-dimensional case) from the Mean Value Theorem. How do you know that $f(x)$ is algebraic? There are plenty of non-algebraic functions whose derivative is not $2x$.

If $f(x)$ is always an integer, it can't be differentiable! (unless it's constant on every interval of its domain) .

Edited the question . I am aware that transcendental function can not be compared with algebraic one. If f(x) produces integer, can it be a transcendental function?

What do you mean "f(x) produces integer"? It makes no sense to talk about the derivative of a function that produces integers for every real value of x. At every point, the derivative will be 0 or undefined.

not for every x (earlier i was in error writing "always" ), f(x) could be fraction but for some x.

Have you noticed that there isn't any part of your question that doesn't have an error?

f(x) could be fraction, meaning rational? Then it still is either constant or non-differentiable.

yes, it was not "defined" well, I apologize.

Sir, consider the case where f(x) is algebraic and differentiable.

Ok

Now what?

The same question ,sir. I have edited the question,please check.

Is f(x) differentiable at all points?

Not every algebraic function is differentiable everywhere

ok

but if, then

can i infer what i said? I would like to know when I can say that .

If two functions f(x) and g(x) are everywhere differentiable and f'(x) != g'(x), then f(x)=g(x) cannot have more than 1 solution. It cannot have 2 solutions. This is a trivial consequence of MVT

then I can say that,ok

You can say that because of MVT, not because of the reason you claimed

Are you denying that this is related to math.stackexchange.com/questions/1214044/… ?

No.

Then why didn't you mention it in your question?

Sir, I am not familiar with the conventions Of SE.

In general, your question should include as much context as possible so that responders understand what you are interested in (and you don't have to rudely tell people not to write about x)

x?

elliptic curves, divisibility

but sir, whenever I ask this question, people would come up with those.

that is not certainly what I was looking for

Sir, I really appreciate this conversation , thanks.