Answer A.
$f(n^2) \text{ versus }f(n)^2 \color{red}{\text{ asymptomatically }}$
1.Take $f(n) = logn$. $ \text{ }f(n^2) = 2logn$ and $f(n)^2 = logn logn$.
Here $\color{green}{f(n^2) < f(n)^2 }\\$
2.Take $f(n) = e^n$. $ \text{ }f(n^2) = e^{n^2}$ and $f(n)^2 = e^n e^n = e^{2n}$.
Here $\color{blue}{f(n^2) > f(n)^2 } $.
You can verify this by taking $log$ both sides in $e^{n^2}$ and $e^{2n}\\$.
3.Take $f(n)$ as polynomial say, $f(n) = n^3$. $f(n^2) = {(n^2)^3 = n^6}$ and $f(n)^2 = {(n^3)^2 = n^6}$.
Here $\mathbf{\color{violet}{f(n^2) = f(n)^2 }}\\ $
Option A.
$f(n^2)=Θ(f(n)^2)$ which is $f(n^2) =f(n)^2$.when f(n) is a polynomial function.
This is TRUE because of point 3.
Option B.
$f(n^2)=o(f(n)^2)$ which is $f(n^2) < f(n)^2$.
Obviously this is false beacuse it depends on $f(n)$.
Option C.
$f(n^2)=O(f(n)^2)$ which is $f(n^2) \leq f(n)^2$. when $f(n)$ is an exponential function.
This is false because of point 2 mentioned above.
Option D
$f(n^2)=Ω(f(n)^2)$ which is $f(n^2) \geq f(n)^2$.
Obviously this is false beacuse it depends on $f(n)$.