The range of $\frac{1}{{\sin }^{2}x+3\sin x\cos x+5{\cos }^{2}x}$ is 

$=\frac{1}{{\sin }^{2}x+3\sin x\cos x+5{\cos }^{2}x}$

$=\frac{1}{1+3\sin x\cos x+4{\cos }^{2}x}$

$=\frac{1}{1+{\displaystyle \frac{3}{2}}\sin 2x+2\left(\cos 2x+1\right)}$

$=\frac{1}{3+{\displaystyle \frac{3}{2}}\sin 2x+2\cos 2x}$

$=\frac{\frac{2}{5}}{\frac{6}{5}+\frac{3}{5}\sin \left(2x\right)+\frac{4}{5}\cos \left(2x\right)}$

Let

$\cos \alpha =\frac{3}{5}$ and $\sin \alpha =\frac{4}{5}$

$f\left(x\right)=\frac{\frac{2}{5}}{\frac{6}{5}+\cos \left(\alpha \right)\sin \left(2x\right)+\sin \left(\alpha \right)\cos \left(2x\right)}$

$f\left(x\right)=\frac{\frac{2}{5}}{\frac{6}{5}+\sin (2x+\alpha )}$

Since $-1\le \sin (2x+\alpha )\le 1$

$\frac{1}{5}\le \sin (2x+\alpha )+\frac{6}{5}\le \frac{11}{5}$

$\frac{5}{11}\le \frac{1}{\sin (2x+\alpha )+\frac{6}{5}}\le 5$

$⟹\frac{2}{11}\le \frac{\frac{2}{5}}{\frac{6}{5}+\sin (2x+\alpha )}\le 2$

$\frac{2}{11}\le f\left(x\right)\le 2$