@ssgvns In option $(A)$, $f(n)$ needs to be polynomial. $\log n$ or $2\log n$ are not polynomial functions.

You can prove the statement of option $(A)$ with the help of limits.

If $\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} \neq 0, \infty$ then $f(n) = \Theta(g(n))$ provided limit exists.

So, consider a real polynomial of degree $m$ as $f(n) = a_0 + a_1n + a_2n^2+ a_3n^3 + a_4n^4+...+a_mn^m$

$f(n^2) = a_0 + a_1n^2 + a_2n^4 + a_3n^6 + a_4n^8 + … + a_m n^{2m}$

Now, $\lim_{n \rightarrow \infty} \left(\frac{(f(n))^2}{f(n^2)} \right) = \lim_{n \rightarrow \infty} \frac{(a_0 + a_1n + a_2n^2+ a_3n^3 + a_4n^4+...+a_mn^m)^2}{a_0 + a_1n^2 + a_2n^4 + a_3n^6 + a_4n^8 + … + a_m n^{2m}} $

$= \lim_{n \rightarrow \infty} \frac{n^2\left(\frac{a_0}{n} + a_1 + a_2n+ a_3n^2 + a_4n^3+...+a_mn^{m-1}\right)^2}{n^2\left(\frac{a_0}{n^2} + a_1 + a_2n^2 + a_3n^4 + a_4n^6 + … + a_m n^{2m-2}\right)}$

$=\lim_{n \rightarrow \infty} \frac{(a_1 + a_2n + a_3n^2+ a_4n^3 +...+a_mn^{m-1})^2}{a_1 + a_2n^2 + a_3n^4 + a_4n^6 + … + a_m n^{2m-2}} $

$= \lim_{n \rightarrow \infty} \frac{n^2\left(\frac{a_1}{n} + a_2 + a_3n+ a_4n^2 + ...+a_mn^{m-2}\right)^2}{n^2\left(\frac{a_1}{n^2} + a_2 + a_3n^2 + a_4n^4 +… + a_m n^{2m-4}\right)}$

$=\lim_{n \rightarrow \infty} \frac{(a_2 + a_3n + a_4n^2+ ...+a_mn^{m-2})^2}{a_2 + a_3n^2 + a_4n^4 + … + a_m n^{2m-4}} $

$= \lim_{n \rightarrow \infty} \frac{n^2\left(\frac{a_2}{n} + a_3 + a_4n+ ...+a_mn^{m-3}\right)^2}{n^2\left(\frac{a_2}{n^2} + a_3 + a_4n^2 + … + a_m n^{2m-6}\right)}$

$=\lim_{n \rightarrow \infty} \frac{(a_3 + a_4n +...+a_mn^{m-3})^2}{a_3 + a_4n^2 + … + a_m n^{2m-6}} $

If we do like this and see the patterm, at the end, we’ll get

$= \lim_{n \rightarrow \infty} \frac{(a_{m-1}+a_m n^{m-(m-1)})^2}{a_{m-1}+a_m n^{2m-(2(m-1))}}$

$= \lim_{n \rightarrow \infty} \frac{(a_{m-1}+a_m n)^2}{a_{m-1}+a_m n^2}$

$= \lim_{n \rightarrow \infty} \frac{n^2\left(\frac{a_{m-1}}{n} + a_m\right)^2}{n^2 \left(\frac{a_{m-1}}{n^2} + a_m\right)}$

$= \frac{a_m^2}{a_m} = a_m$

Since, $a_m \neq 0, \infty$, It means $f(n)^2 = \Theta(f(n^2))$ or $f(n^2) = \Theta (f(n)^2)$

(You can also see that $(f(n))^2$ is a polynomial in $n$ with highest exponent as $2m$, so, $f(n)^2 = \Theta(n^{2m})$ and $f(n^2)$ is also a polynomial in $n$ with highest exponent as $2m$ because $n$ is replaced with $n^2$, so $f(n^2) = \Theta(n^{2m})$ and so, $f(n^2) = \Theta(f(n)^2)$)