Given function:
f(x)=e^{-x}\cos(-x)-\frac{\sin x}{\pi-e^x}
f(x)=e^{-x}\cosx+\frac{\sin x}{e^x-\pi}
Differentiating above function w.r.t. x using chain rule & quotient rule as follows
f'(x)=d/dx(e^{-x}\cosx+\frac{\sin x}{e^x-\pi})
=e^{-x}(-\sinx)+\cos x(-e^{-x})+\frac{(e^x-\pi)\cos x-\sin x(e^x)}{(e^x-\pi)^2}
=-e^{-x}(\sinx+\cos x)+\frac{e^x(\cos x-\sin x)-\pi\cos x}{(e^x-\pi)^2}
Setting x=\pi/6 in above expression we get
f'(\pi/6)
=-e^{-\pi/6}(\sin(\pi/6)+\cos (\pi/6))+\frac{e^{\pi/6}(\cos (\pi/6)-\sin (\pi/6))-\pi\cos (\pi/6)}{(e^{\pi/6}-\pi)^2}
=-0.80921-0.9953
=-1.8045
Since f'(\pi/6)<0 hence the given function f(x) is decreasing at x=\pi/6