The derivative of $f(x)$ is $\frac{8(3\cos^2 x+3\sin^2 x-10 \sin x)}{(3 \sin x+4 \cos x-10)^2}$. Note that the maximum of $f(x)$ occurs when $f'(x)=0$. Thus we need $3\cos^2 x+3\sin^2 x-10 \sin x=0 \Rightarrow 3\cos^2 x+3\sin^2 x=10 \sin x$. The LHS is equal to $3(1)$ by the Pythagorean identity, so $\sin(x)=\frac{3}{10}$ and therefore, $\cos(x)=\frac{\sqrt{91}}{10}$. Hence, the maximum is $\boxed{\frac{107+8\sqrt{91}}{75}}$.

No need for calculus Using diff of squares the equation becomes $(3\sin{x}-10)^2-16\cos^2{x}$ Now use $\cos^2{x}=1-\sin^2{x}$ u get quadratic in $\sin$ rest is simple

wait oops... 1) thought it was $f(x)=\frac{3 \sin x -4 \cos x - 10}{3 \sin x +4 \cos x -10}$ sigh... 2) found maximum, not minimum -_- I'll edit my solution tomorrow since it's getting lateee