Find the smallest value of $f(x) := \left({1\over9}+{32\over \sin(x)}\right)\left({1\over32}+{9\over \cos(x)}\right)$ on the interval $(0,\pi/2)$

Since both $\cos$ and $\sin $ are positive in $(o,{\pi\over 2})$ we can use Cauchy inequaliy:

$$ (a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$$

$$\bigg({1\over9}+{32\over \sin(x)}\bigg)\bigg({1\over32}+{9\over \cos(x)}\bigg)\geq \bigg({1\over \sqrt{288}}+{\sqrt{288}\over \sqrt{\sin(x)\cos(x)}}\bigg)^2\geq \bigg({1\over 12\sqrt{2}}+24\bigg)^2$$

We used here $$\sin(x)\cos(x)= {1\over 2}\sin (2x) \leq {1\over 2}$$ with equality at $x={\pi \over 4}$. So $$y_{min} = \bigg({1\over 12\sqrt{2}}+24\bigg)^2$$

$$1+\frac{288}{\sin x}$$ is a decreasing function in $\left(0,\dfrac\pi2\right)$ and its symmetric

$$1+\frac{288}{\cos x}$$ is increasing.

Hence the minimum occurs ar $x=\dfrac\pi4$.

Here's how to continue with your idea:

Let $\sin{x}+\cos{x}=t$.

Then, by the Cauchy-Schwartz inequality we have $$1<t=\sin{x}+\cos{x}\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2,$$ where the equality occurs for $x=\frac{\pi}{4},$ and since $$\left(\frac{t+288}{t^2-1}\right)'=-\frac{t^2+576t+1}{(t^2-1)^2}<0,$$ we obtain $$ f(x) =\frac{1}{288}+\frac{2(t+288)}{t^2-1} \ge \frac{1}{288}+\frac{2(\sqrt{2}+288)}{2-1} =576 + \frac{1}{288} + 2\sqrt{2}$$ and we are done!