Monotonicity of trigonometric function

You have $\sin2x=2\sin x \cos x$ and $\sin 3x=\sin x \cos 2x + \cos x \sin 2x =\sin x (\cos 2 x + 2 \cos^2 x)$, and similar formulas for higher $\sin (kx)$. Then $$\frac{\sin(3x)}{\sin x} = \cos 2 x + 2\cos^2x $$ where the RHS is decreasing in $x$ and positive as long as $\cos 2x>0$. That gives you an interval where $\frac{\sin x}{\sin 3x}$ is increasing.

Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).

$$\begin{aligned}\frac{\sin(x)}{\sin((k+1)x)}&=\frac{\sin(x)}{\sin(kx)\cos(x)+\cos(kx)\sin(x)}\\ &=\frac{1}{\frac{\sin(kx)}{\sin(x)}\cos(x)+\cos(kx)} \end{aligned}$$

Hence: $$\begin{aligned}\frac{\sin(x)}{\sin(kx)}\text{ increasing}&\iff\frac{\sin(kx)}{\sin(x)}\text{ decreasing}\\ &\Rightarrow \frac{\sin(kx)}{\sin(x)}\cos(x)\text{ decreasing}\\ &\iff\frac{\sin(kx)}{\sin(x)}\cos(x)+\cos(kx)\text{ decreasing}\\ &\iff\frac1{\frac{\sin(kx)}{\sin(x)}\cos(x)+\cos(kx)}\text{ increasing}\\ \\ &\iff\frac{\sin(x)}{\sin((k+1)x)}\text{ increasing}\\ \end{aligned}$$

Let $P(k)$ denote the proposition that $\frac{\sin(x)}{\sin(kx)}$ is increasing. Hence, we've shown that $P(k)\rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $\frac{\sin(x)}{\sin((2k+1)x)}$ is increasing over $[0,\frac{\pi}{2(2k+1)}]$ for all $k\in\mathbb{N}$.