The conjugate factor theorem states that for a polynomial $p(x)$ with real coefficients, the complex roots come in conjugate pairs, or if $a+bi$ is a root, then $a-bi$ is also a root. From this we see that your polynomial has the roots $$\frac{2}{3}, -1, 3+\sqrt{2}i, 3-\sqrt{2}i.$$ Therefore, $$(x-\frac{2}{3})(x+1)(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0.$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x^2-6x+11)=0.$$ Continuing the expansion results in the polynomial $$p(x)=\frac{1}{3}(3x^4-17x^3+25x^2+23x-22).$$
First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.
There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.
Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.
Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.
Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.
Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)."
The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.
The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.
Best Answer
What you are asking for is a direct consequence of Rolle's Theorem. Basically Rolle's Theorem states that if a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a) = f(b)$, then $\exists c \in (a,b)$ such that $f'(c) = 0$. If you apply Rolle's Theorem to the desired polynomial between any two real roots of $f$, (a polynomial is continuous and differentiable), you will get what you need directly.
For more information on Rolle's Theorem
As @DonAntonio and @Omnomnomnom have pointed out, for repeated roots, the zero will be a root of the polynomial and it's derivatives as well so this can be considered separately.