Even, Odd is Ordinary

We have, $\large e(x)=f(x)-o(x) ...(1)$ replacing $x$ with $-x,$ $\large e(x)=f(-x)+o(x) ...(2)$ Further it was given, $\large e(x)=o(x)-x^2 ...(3)$ comparing $(2)$ and $(3)$ we get $\large f(-x)=-x^2$ $\Rightarrow f(2)=-4$

Substitute 2 into the second equation to get

$e(2)+4=o(2)\dots(3)$

Substitute -2 into the second equation and use the properties of even and odd functions to get

$e(2)+4=-o(2)\dots(4)$

Add equations (3) and (4) to see that

$e(2)+4=0$

$\implies e(2)=-4\dots(5)$

equations (3) and(5) give

$o(2)=0\dots(6)$

Substituting (5) and (6) into the first equation in the problem gives

$f(2)=-4$

$o(x)=e(x)+{ x }^{ 2 }\\ \\ Substituting\quad x\quad by\quad -x.\\ \\ o(-x)=e(-x)+{ (-x) }^{ 2 }\\ o(-x)=e(x)+{ x }^{ 2 }\\ So\quad o(-x)=o(x)\qquad \qquad \qquad \qquad (1)\\ \\ We\quad knew\quad that\\ \quad o(-x)=-o(x)\quad \qquad \qquad \qquad \qquad (2)\\ Using\quad (1)\quad and\quad (2),\quad we\quad get\\ \\ -o(x)=o(x)\\ So\quad o(x)=0\\ \\ So\quad e(x)+{ x }^{ 2 }=0\\ So\quad e(x)=-{ x }^{ 2 }\\ So\quad f(2)=e(2)+o(2)=\boxed { 4 }$

I found the biggest clue in the second equation

$e(x) + x^2 = o(x)$

Here we are adding two even functions to get an odd function. We know if we add two even function, the result MUST be even function.

Thus $o(x)$ must be an even function. But the problem says that $o(x)$ is an odd function. The only function that is both odd as well as even is: $o(x) = 0$

Thus, from the first equation we get: $e(x) = f(x)$

And from the second one we get: $e(x) = -x^2$

Combining, we have:

$f(x) = -x^2$ or, $f(2) = -4$