A second of thinking tells you that no function achieves that !
$$\sqrt{x+y}\ne\sqrt x+\sqrt y,\frac1{x+y}\ne\frac1x+\frac1y,\log(x+y)\ne\log x+\log x,\cdots$$
$$\frac xy+1\ne\frac{x+1}{y+1},\tan\frac xy\ne\frac{\tan x}{\tan y}\cdots$$
There are just two exceptions:
$$a(x+y)=ax+ay$$
and
$$\left(\frac xy\right)^a=\frac{x^a}{y^a}.$$
So you'd better ask why the linear function is additive and why the power function is multiplicative.
If you want to find all additive functions, i.e. such that
$$f(x+y)=f(x)+f(y),$$ you immediately see that
$$f(2x)=2f(x)$$ and by induction
$$f(nx)=nf(x).$$
This generalizes to rationals,
$$qf\left(\frac pqx\right)=qpf\left(\frac xq\right)=pf(x)\implies f\left(\frac pqx\right)=\frac pqf(x),$$ and to reals
$$f(rx)=rf(x),$$ but the proof is more technical.
Now, setting $r\to x,x\to1$,
$$f(x)=f(1)\,x=ax.$$
For the multiplicative functions
$$g(xy)=g(x)g(y)$$
consider the function
$$f(x):=\log g(e^x)$$ and observe that it is additive, so that
$$f(x)=\log g(e^x)=ax,$$
$$g(e^x)=e^{ax},$$
$$g(x)=x^a.$$
Best Answer
Pure definition:
A function $f$ is odd if $f(-x) = - f(x)$ for all $x$.
So $\sin((-x)^3) = \sin(-x^3) = - \sin (x^3)$.
So it is odd.
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One thing to note, is that if $f$ is odd, and $g$ is also odd then $f\circ g$ is odd because:
$f(g(-x) = f(-g(x)) = -f(g(x))$.
And $x^3$ is odd because $(-x)^3 = (-1*x)^3 = (-1)^3x^3 = (-1)*x^3 = -x^3$
An $\sin x$ is odd because .... well, it's a basic trigonometric identity that $\sin (-x) = -\sin (x)$. Just draw a picture. The $y$ value of a point of a circle at an angle will be the negative of the $y$ value of the angle in the opposite direction.
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Only if you can also say that $-u = (-x)^3$.