Not simultaneously even & odd function.

JEE style.

We can easily observe that $f(x) =x$ satisfies all given conditions.

Firstly I feel that this does not look like a calculus question at all, but more of a functional equation problem. Just observing $f(x)$ is not sufficient to solve this problem, but to prove it.

Disclaimer: The problems states the derivative exists at $x=0$ and asks for $f'(10)$ . So I assume $f$ is continuous (which I think should be added to the question in some subtle way like $f'(x)\ge 0$ ).

Let $y=0$ , then $f(x)=f(x)+f(0)^{2n+1} \iff f(0)=0$ . Let $x=0$ , $f(y^{2n+1})=f(y)^{2n+1}$ . Hence substituting that into the original equation and let $y^{2n+1}=z$ , and note that $z$ covers all reals. Hence we get the Cauchy equation $f(x+z)=f(x)+f(z)$ . Note that this and $f(y^{2n+1})=f(y)^{2n+1}$ can imply the original equation, so the original equation is equivalent to these two equations. So we know the solution to Cauchy equation is $f(x)=ax, a\in \mathbb{R}$ , since $f$ is not zero, hence $a$ is nonzero. Now for any $y$ , $ay^{2n+1}=f(y^{2n+1})=f(y)^{2n+1}=(ay)^{2n+1}$ which forces $a=1$ as $f'(0)\ge 0$ which is the only application of calculus here. Hence $f'(10)=1$ .

Note that if we did not use the assumption that $f$ is continuous, I couldn't see how to make more progress.

As it given that f(x) is differentiable .. we can take partial differential wrt x anf clearly sees that f'(x)=constant .. the solution to functional equation is a line... and satisfying conditions f'(x)=1