Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$
Note that by denoting $f(x) = \tan x \sin x -x^2$, you found that $$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x) = \sin x (1+3\sec^2 x)(2\sec^2 x -1 ) \geq 0 $$ Hence $f''(x)$ is increasing, with $f''(0)=0$, we conclude that $f''(x) \geq 0$.
Hence $f'(x)$ is increasing, with $f'(0)=0$, we conclude that $f'(x) \geq 0$.
Hence $f(x)$ is increasing, with $f(0)=0$, we conclude that $f(x) \geq 0$.
This is what we wish to prove.
From a more advanced perspective, the inequality follows from the fact that Taylor expansion of $$\tan x \sin x = x^2+\frac{x^4}{6}+\cdots$$ at $x=0$ have all coefficients positive, the radius of convergence of this series is $\pi/2$.
To see why all coefficients are positive, write $$\tan x \sin x = \frac{1}{\cos x} - \cos x$$
The Taylor expansion of $\sec x$ at $x=0$ is $$\sec x = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!} x^{2n}$$ where $E_{2n}$ are Euler number. The fact that $(-1)^n E_{2n}$ is positive follows from the series evaluation: $$\beta(2n+1) = \frac{(-1)^n E_{2n} \pi^{2n+1}}{4^{2n+1} (2n)!}$$ with $\beta(n)$ the Dirichlet beta function.
Also note that we have $|E_{2n}| > 1 $ when $n>1$, hence the power series of $\frac{1}{\cos x}-\cos x$ has all coefficients positive.
From this, you might want to prove the stronger inequality:
When $0<x<\frac{\pi}{2}$, $$\tan x \sin x > x^2 + \frac{x^4}{6} $$ $$\tan x \sin x > x^2 + \frac{x^4}{6} + \frac{31x^6}{360} $$
I believe the simplest proof is through the Cauchy-Schwarz inequality:
$$\tan(x)\sin(x)=\int_{0}^{x}\frac{d\theta}{\cos^2\theta}\int_{0}^{x}\cos(\theta)\,d\theta\geq\left(\int_{0}^{x}\frac{d\theta}{\sqrt{\cos\theta}}\right)^2\geq\left(\int_{0}^{x}d\theta\right)^2=x^2. $$ In a similar fashion, for any $x\in\left(0,\frac{\pi}{2}\right)$ we have $\frac{\tan x }{x}\geq\left(\frac{x}{\sin x}\right)^2$ by Holder's inequality.
We need to prove that $$\frac{\sin^2x}{\cos{x}}>x^2$$ or $f(x)>0$, where $$f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x.$$ Now, let $\cos{x}=t^2$, where $0<t<1$.
Thus, $$f'(x)=\frac{1+\cos^2x}{2\sqrt{\cos^3x}}-1=\frac{(1-t)(1+t+t^2-t^3)}{2t^3}>0,$$ which says $f(x)>f(0)=0$ and we are done!