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Examine whether $f(x)=3+(x-3)^\frac23$ has an extremum point at $x=3$

I know that if $c$ be the extreme point then $f'(c)=0$. So we performed $f'(3)=0$ but I unable to solve this.

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    @HarshKumar That ambiguity is why I asked the OP to check the equation. – 2017-02-12
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    Both those $3$'s are irrelevant to the real problem, which is: what happens to the function $f(x)=x^{2/3}$ at $x=0$? See if you can draw a graph. It should show you what's going on. – 2017-02-12
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    And to do that, you will need to decide what your definition of $x^{2/3}$ is when $x<0$ (which is all the question is about, actually). – 2017-02-12
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    @Did: I would say that is unambiguous here. You can square first, and then take the cube root; or take the cube root first, and then square. How else might you sensibly define it? – 2017-02-12
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    @TonyK Dunno, but the point is that there is no canonical definition here. – 2017-02-12
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    @Did: There is an _obvious_ approach, and no other approach makes sense. Why pick holes in that? – 2017-02-12
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    @TonyK Because one routinely sees quantities $x^a$ manipulated carelessly on this site, even when $x<0$ and $a$ is irrational, say. But you are probably right that the rewriting in egreg's answer shows there is no problem in the present case. – 2017-02-12

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It is not true that $f'(c)=0$ at extrema in general. If the function is differentiable at $c$ and $c$ is an interior point of the domain and it is an extremum, then $f'(c)=0$.

I assume you mean $$ f(x)=3+\bigl((x-3)^2\bigr)^{1/3} $$ It has an absolute minimum at $3$, because $f(3)=3$ and $f(x)>3$ if $x\ne3$.

The criterion with the derivative cannot be exploited here.

The function is everywhere defined and its derivative is $$ f'(x)=-\frac{1}{\sqrt[3]{3-x}} $$ which is undefined at $3$ and, indeed, the function is not differentiable at $3$.

The function is continuous at $3$, so we can use l’Hôpital’s theorem for computing the limit $$ \lim_{x\to3^+}\frac{f(x)-f(3)}{x-3}=\lim_{x\to3^+}f'(x)=-\infty $$ Thus the function is not differentiable at $3$.

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    "and, indeed, the function is not differentiable at 3" Something that you state as a fact but did not prove, right? – 2017-02-12
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    @Did Yes, I left the proof to the reader; but, since the limit of the derivative at $3$ is $\pm\infty$ (depending on what side you approach $3$), l’Hôpital's theorem says that the function is not differentiable. – 2017-02-12
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    Seeing how users regularly err on this point (even on this very page), it might be wiser, either to be fully explicit about it, or to avoid mentioning the argument altogether. – 2017-02-12
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    @Did I reworded my answer. – 2017-02-12
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    Yes, with the inconvenience that now you are appealing to a version of L'H (the one where one does not assume differentiability at the point of interest) which is more elaborate than the result to be shown... Oh well. – 2017-02-12
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    @Did Just a convenience; computing the limit by rationalization is easy, but very tedious and not general. Anyway, it is not essential for discussing whether $3$ is an extremum. – 2017-02-12
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    "Anyway, it is not essential for discussing whether 3 is an extremum." On this we can fully agree. – 2017-02-12
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    Thanks to all of you for helps me – 2017-02-12