Let $y=\frac{{\sin }^{3}x}{\cos x}+\frac{{\cos }^{3}x}{\sin x}$  where $0<x<\frac{\pi }{2}$ , then the minimum value of y is 

$y=\frac{1-2{\sin }^{2}x{\cos }^{2}x}{\sin x\cos x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}-2\sin x\cos x=$

$(\tan x+\cot x)-\sin 2x$

Now,  $(\tan x+\cot x)$ is minimum at  $x=\frac{\pi }{4}$ and $\sin 2x$ is maximum at 

$x=\frac{\pi }{4}$ 
$\therefore y_{\min }$  occurs at  $x=\frac{\pi }{4}and{y}_{\min }=1$